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== 1.10#h(0.6em)Exercises


    These are homework exercises to accompany Libl's "#link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)")[Differential Equations for Engineering]" Textmap. This is a textbook targeted for a one semester first course on differential equations, aimed at engineering students. Prerequisite for the course is the basic calculus sequence.


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.1%3A_Integrals_as_solutions")[1.1: Integrals as solutions]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve for #math.equation(block: false, alt: "v")[$v$], and then solve for #math.equation(block: false, alt: "x")[$x$]. Find #math.equation(block: false, alt: "x open parenthesis 10 close parenthesis")[$x ( 10 )$] to answer the question.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals x squared plus x")[$frac(d y, d x) = x^(2) + x$] for #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 3")[$y ( 1 ) = 3$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals sin open parenthesis 5 x close parenthesis")[$frac(d y, d x) = sin ( 5 x )$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 2")[$y ( 0 ) = 2$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals the fraction 1 over x squared minus 1")[$frac(d y, d x) = frac(1, x^(2) − 1)$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals y cubed")[$y^(′) = y^(3)$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(little harder)]


      Solve #math.equation(block: false, alt: "y prime equals open parenthesis y minus 1 close parenthesis open parenthesis y plus 1 close parenthesis")[$y^(′) = ( y − 1 ) ( y + 1 )$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 3")[$y ( 0 ) = 3$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals the fraction 1 over y plus 1")[$frac(d y, d x) = frac(1, y + 1)$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(harder)]


      Solve #math.equation(block: false, alt: "y double prime equals sin x")[$y^(″) = sin x$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$], #math.equation(block: false, alt: "y prime open parenthesis 0 close parenthesis equals 2")[$y^(′) ( 0 ) = 2$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A spaceship is traveling at the speed #math.equation(block: false, alt: "2 t squared plus 1")[$2 t^(2) + 1$] #super[km]/#sub[s] (#math.equation(block: false, alt: "t")[$t$] is time in seconds). It is pointing directly away from earth and at time #math.equation(block: false, alt: "t equals 0")[$t = 0$] it is 1000 kilometers from earth. How far from earth is it at one minute from time #math.equation(block: false, alt: "t equals 0")[$t = 0$]?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d x over d t equals sin open parenthesis t squared close parenthesis plus t")[$frac(d x, d t) = sin ( t^(2) ) + t$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 20")[$x ( 0 ) = 20$]. It is OK to leave your answer as a definite integral.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A dropped ball accelerates downwards at a constant rate #math.equation(block: false, alt: "9.8")[$9.8$] meters per second squared. Set up the differential equation for the height above ground #math.equation(block: false, alt: "h")[$h$] in meters. Then supposing #math.equation(block: false, alt: "h open parenthesis 0 close parenthesis equals 100 meters")[$h ( 0 ) = 100 " meters"$], how long does it take for the ball to hit the ground.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the general solution of #math.equation(block: false, alt: "y prime equals e to the power x")[$y^(′) = e^(x)$], and then #math.equation(block: false, alt: "y prime equals e to the power y")[$y^(′) = e^(y)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals e to the power x plus x")[$frac(d y, d x) = e^(x) + x$] and #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 10")[$y ( 0 ) = 10$].


      Answer


      #math.equation(block: true, alt: "y equals e to the power x plus the fraction x squared over 2 plus 9")[$y = e^(x) + frac(x^(2), 2) + 9$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x prime equals the fraction 1 over x squared")[$x^(′) = frac(1, x^(2))$], #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis equals 1")[$x ( 1 ) = 1$].


      Answer


      #math.equation(block: true, alt: "x equals open parenthesis 3 t minus 2 close parenthesis to the power 1 / 3")[$x = ( 3 t − 2 )^(1 / 3)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x prime equals the fraction 1 over cos open parenthesis x close parenthesis")[$x^(′) = frac(1, cos ( x ))$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals the fraction π over 2")[$x ( 0 ) = frac(π, 2)$].


      Answer


      A#math.equation(block: false, alt: "x equals sin to the power minus 1 open parenthesis t plus 1 close parenthesis")[$x = sin^(− 1) ( t + 1 )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Sid is in a car traveling at speed #math.equation(block: false, alt: "10 t plus 70")[$10 t + 70$] miles per hour away from Las Vegas, where #math.equation(block: false, alt: "t")[$t$] is in hours. At #math.equation(block: false, alt: "t equals 0")[$t = 0$] the Sid is 10 miles away from Vegas. How far from Vegas is Sid #math.equation(block: false, alt: "2")[$2$] hours later?


      Answer


      #math.equation(block: true, alt: "170")[$170$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals y double prime")[$y^(′) = y^(″)$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$], where #math.equation(block: false, alt: "n")[$n$] is a positive integer. Hint: You have to consider different cases.


      Answer


      If #math.equation(block: false, alt: "n not equal to 1")[$n ≠ 1$], then #math.equation(block: false, alt: "y equals open parenthesis open parenthesis 1 minus n close parenthesis x plus 1 close parenthesis to the power 1 / open parenthesis 1 minus n close parenthesis")[$y = ( ( 1 − n ) x + 1 )^(1 / ( 1 − n ))$]. If #math.equation(block: false, alt: "n equals 1")[$n = 1$], then #math.equation(block: false, alt: "y equals e to the power x")[$y = e^(x)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The rate of change of the volume of a snowball that is melting is proportional to the surface area of the snowball. Suppose the snowball is perfectly spherical. The volume (in centimeters cubed) of a ball of radius #math.equation(block: false, alt: "r")[$r$] centimeters is #math.equation(block: false, alt: "the fraction 4 over 3 close parenthesis π r cubed")[$frac(4, 3) \) π r^(3)$]. The surface area is #math.equation(block: false, alt: "4 π r squared")[$4 π r^(2)$]. Set up the differential equation for how the radius #math.equation(block: false, alt: "r")[$r$] is changing. Then, suppose that at time #math.equation(block: false, alt: "t equals 0")[$t = 0$] minutes, the radius is #math.equation(block: false, alt: "10")[$10$] centimeters. After #math.equation(block: false, alt: "5")[$5$] minutes, the radius is #math.equation(block: false, alt: "8")[$8$] centimeters. At what time #math.equation(block: false, alt: "t")[$t$] will the snowball be completely melted?


      Answer


      The equation is #math.equation(block: false, alt: "r prime equals minus C")[$r^(′) = − C$] for some constant #math.equation(block: false, alt: "C")[$C$]. The snowball will be completely melted in #math.equation(block: false, alt: "25")[$25$] minutes from time #math.equation(block: false, alt: "t equals 0")[$t = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the general solution to #math.equation(block: false, alt: "y to the power ⁗ equals 0")[$y^(⁗) = 0$]. How many distinct constants do you need?


      Answer


      #math.equation(block: false, alt: "y equals A x cubed plus B x squared plus C x plus D")[$y = A x^(3) + B x^(2) + C x + D$], so #math.equation(block: false, alt: "4")[$4$] constants.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.2%3A_Slope_fields")[1.2: Slope fields]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Sketch slope field for #math.equation(block: false, alt: "y prime equals e to the power x minus y")[$y^(′) = e^(x − y)$]. How do the solutions behave as #math.equation(block: false, alt: "x")[$x$] grows? Can you guess a particular solution by looking at the slope field?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Sketch slope field for #math.equation(block: false, alt: "y prime equals x squared")[$y^(′) = x^(2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Sketch slope field for #math.equation(block: false, alt: "y prime equals y squared")[$y^(′) = y^(2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Is it possible to solve the equation #math.equation(block: false, alt: "y prime equals the fraction x y over cos x")[$y^(′) = frac(x y, cos x)$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$]? Justify.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Is it possible to solve the equation #math.equation(block: false, alt: "y prime equals y the square root of | x |")[$y^(′) = y sqrt(| x |)$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$]? Is the solution unique? Justify.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Match equations #math.equation(block: false, alt: "y prime equals 1 minus x")[$y^(′) = 1 − x$], #math.equation(block: false, alt: "y prime equals x minus 2 y")[$y^(′) = x − 2 y$], #math.equation(block: false, alt: "y prime equals x open parenthesis 1 minus y close parenthesis")[$y^(′) = x ( 1 − y )$] to slope fields. Justify.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Take #math.equation(block: false, alt: "y prime equals f open parenthesis x , y close parenthesis")[$y^(′) = f ( x , y )$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$], where #math.equation(block: false, alt: "f open parenthesis x , y close parenthesis greater than 1")[$f ( x , y ) > 1$] for all #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$]. If the solution exists for all #math.equation(block: false, alt: "x")[$x$], can you say what happens to #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$] as #math.equation(block: false, alt: "x")[$x$] goes to positive infinity? Explain.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Take #math.equation(block: false, alt: "open parenthesis y minus x close parenthesis y prime equals 0")[$( y − x ) y^(′) = 0$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$].


      + Find two distinct solutions.
+ Explain why this does not violate Picard’s theorem.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "y prime equals f open parenthesis x , y close parenthesis")[$y^(′) = f ( x , y )$]. What will the slope field look like, explain and sketch an example, if you know the following about #math.equation(block: false, alt: "f open parenthesis x , y close parenthesis")[$f ( x , y )$]:


      + #math.equation(block: false, alt: "f")[$f$] does not depend on #math.equation(block: false, alt: "y")[$y$].
+ #math.equation(block: false, alt: "f")[$f$] does not depend on #math.equation(block: false, alt: "x")[$x$].
+ #math.equation(block: false, alt: "f , open parenthesis t , t close parenthesis equals 0")[$f , ( t , t ) = 0$] for any number #math.equation(block: false, alt: "t")[$t$].
+ #math.equation(block: false, alt: "f open parenthesis x , 0 close parenthesis equals 0")[$f ( x , 0 ) = 0$] and #math.equation(block: false, alt: "f open parenthesis x , 1 close parenthesis equals 1")[$f ( x , 1 ) = 1$] for all #math.equation(block: false, alt: "x")[$x$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find a solution to #math.equation(block: false, alt: "y prime equals | y |")[$y^(′) = | y |$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$]. Does Picard’s theorem apply?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take an equation #math.equation(block: false, alt: "y prime equals open parenthesis y minus 2 x close parenthesis g open parenthesis x , y close parenthesis plus 2")[$y^(′) = ( y − 2 x ) g ( x , y ) + 2$] for some function #math.equation(block: false, alt: "g open parenthesis x , y close parenthesis")[$g ( x , y )$]. Can you solve the problem for the initial condition #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$], and if so what is the solution?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(challenging)]


      Suppose #math.equation(block: false, alt: "y prime equals f open parenthesis x , y close parenthesis")[$y^(′) = f ( x , y )$] is such that #math.equation(block: false, alt: "f open parenthesis x , 1 close parenthesis equals 0")[$f ( x , 1 ) = 0$] for every #math.equation(block: false, alt: "x , f")[$x , f$] is continuous and #math.equation(block: false, alt: "the fraction ∂ f over ∂ y")[$frac(∂ f, ∂ y)$] exists and is continuous for every #math.equation(block: false, alt: "x")[$x$] and #math.equation(block: false, alt: "y")[$y$].


      + Guess a solution given the initial condition #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
+ Can graphs of two solutions of the equation for different initial conditions ever intersect?
+ Given #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$], what can you say about the solution. In particular, can #math.equation(block: false, alt: "y open parenthesis x close parenthesis greater than 1")[$y ( x ) > 1$] for any #math.equation(block: false, alt: "x")[$x$]? Can #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals 1")[$y ( x ) = 1$] for any #math.equation(block: false, alt: "x")[$x$]? Why or why not?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Sketch the slope field of #math.equation(block: false, alt: "y prime equals y cubed")[$y^(′) = y^(3)$]. Can you visually find the solution that satisfies #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$]?


      Answer


      #figure(figph[Slope field of blue dashes depending only on height, with an equilibrium row of horizontal dashes across the middle; dashes tilt up to the right above it and down to the right below, carrying solutions away.], alt: "Slope field of blue dashes depending only on height, with an equilibrium row of horizontal dashes across the middle; dashes tilt up to the right above it and down to the right below, carrying solutions away.", caption: none)


      #math.equation(block: false, alt: "y equals 0")[$y = 0$] is a solution such that #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Is it possible to solve #math.equation(block: false, alt: "y prime equals x y")[$y^(′) = x y$] for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$]? Is the solution unique?


      Answer


      Yes a solution exists. The equation is #math.equation(block: false, alt: "y prime equals f open parenthesis x , y close parenthesis")[$y^(′) = f ( x , y )$] where #math.equation(block: false, alt: "f open parenthesis x , y close parenthesis equals x y")[$f ( x , y ) = x y$]. The function #math.equation(block: false, alt: "f open parenthesis x , y close parenthesis")[$f ( x , y )$] is continuous and #math.equation(block: false, alt: "the fraction ∂ f over ∂ y equals x")[$frac(∂ f, ∂ y) = x$], which is also continuous near #math.equation(block: false, alt: "open parenthesis 0 , 0 close parenthesis")[$( 0 , 0 )$]. So a solution exists and is unique. (In fact, #math.equation(block: false, alt: "y equals 0")[$y = 0$] is the solution.)
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Is it possible to solve #math.equation(block: false, alt: "y prime equals the fraction x over x squared minus 1")[$y^(′) = frac(x, x^(2) − 1)$] for #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 0")[$y ( 1 ) = 0$]?


      Answer


      No, the equation is not defined at #math.equation(block: false, alt: "open parenthesis x , y close parenthesis equals open parenthesis 1 , 0 close parenthesis")[$( x , y ) = ( 1 , 0 )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Match equations #math.equation(block: false, alt: "y prime equals sin x")[$y^(′) = sin x$], #math.equation(block: false, alt: "y prime equals cos y")[$y^(′) = cos y$], #math.equation(block: false, alt: "y prime equals y cos open parenthesis x close parenthesis")[$y^(′) = y cos ( x )$] to slope fields. Justify.


      
      Answer


      + #math.equation(block: false, alt: "y prime equals cos y")[$y^(′) = cos y$]
+ #math.equation(block: false, alt: "y prime equals y cos open parenthesis x close parenthesis")[$y^(′) = y cos ( x )$]
+ #math.equation(block: false, alt: "y prime equals sin x")[$y^(′) = sin x$]


      Justification left to reader.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
#emph[(tricky)]


      Suppose


      #math.equation(block: true, alt: "0 i f y greater than 0 ,; 1 i f y less than or equal to 0 .")[$0 i f #h(0.222em) y > 0 , \ 1 i f #h(0.222em) y ≤ 0 .$]


      Does #math.equation(block: false, alt: "y prime equals f open parenthesis y close parenthesis")[$y^(′) = f ( y )$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$] have a continuously differentiable solution? Does Picard apply? Why, or why not?


      Answer


      Picard does not apply as #math.equation(block: false, alt: "f")[$f$] is not continuous at #math.equation(block: false, alt: "y equals 0")[$y = 0$]. The equation does not have a continuously differentiable solution. Suppose it did. Notice that #math.equation(block: false, alt: "y prime open parenthesis 0 close parenthesis equals 1")[$y^(′) ( 0 ) = 1$]. By the first derivative test, #math.equation(block: false, alt: "y open parenthesis x close parenthesis greater than 0")[$y ( x ) > 0$] for small positive #math.equation(block: false, alt: "x")[$x$]. But then for those #math.equation(block: false, alt: "x")[$x$] we would have #math.equation(block: false, alt: "y prime open parenthesis x close parenthesis equals 0")[$y^(′) ( x ) = 0$], so clearly the derivative cannot be continuous.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider an equation of the form #math.equation(block: false, alt: "y prime equals f open parenthesis x close parenthesis")[$y^(′) = f ( x )$] for some continuous function #math.equation(block: false, alt: "f")[$f$], and an initial condition #math.equation(block: false, alt: "y open parenthesis x sub 0 close parenthesis equals y sub 0")[$y ( x_(0) ) = y_(0)$]. Does a solution exist for all #math.equation(block: false, alt: "x")[$x$]? Why or why not?


      Answer


      The solution is #math.equation(block: false, alt: "y open parenthesis x close parenthesis equals ∫ x sub 0 x f open parenthesis s close parenthesis d s plus y sub 0")[$y ( x ) = ∫_(x_(0))^(x) f ( s ) d s + y_(0)$], and this does indeed exist for every #math.equation(block: false, alt: "x")[$x$].
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.3%3A_Separable_Equations")[1.3: Separable Equations]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals the fraction x over y")[$y^(′) = frac(x, y)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals x squared y")[$y^(′) = x^(2) y$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d x over d t equals open parenthesis x squared minus 1 close parenthesis")[$frac(d x, d t) = ( x^(2) − 1 )$], for #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d x over d t equals x sin open parenthesis t close parenthesis")[$frac(d x, d t) = x sin ( t )$], for #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1")[$x ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals x y plus x plus y plus 1")[$frac(d y, d x) = x y + x + y + 1$]. Hint: Factor the right hand side.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x y prime equals y plus 2 x squared y")[$x y^(′) = y + 2 x^(2) y$], where #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 1")[$y ( 1 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction d y over d x equals the fraction y squared plus 1 over x squared plus 1")[$frac(d y, d x) = frac(y^(2) + 1, x^(2) + 1)$], for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an implicit solution for #math.equation(block: false, alt: "the fraction d y over d x equals the fraction x squared plus 1 over y squared plus 1")[$frac(d y, d x) = frac(x^(2) + 1, y^(2) + 1)$], for #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an explicit solution for #math.equation(block: false, alt: "y prime equals x e to the power minus y")[$y^(′) = x e^(− y)$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an explicit solution for #math.equation(block: false, alt: "x y prime equals e to the power minus y")[$x y^(′) = e^(− y)$], for #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 1")[$y ( 1 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an explicit solution for #math.equation(block: false, alt: "y prime equals y e to the power minus x squared")[$y^(′) = y e^(− x^(2))$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$]. It is alright to leave a definite integral in your answer.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose a cup of coffee is at 100 degrees Celsius at time #math.equation(block: false, alt: "t equals 0")[$t = 0$], it is at 70 degrees at #math.equation(block: false, alt: "t equals 10")[$t = 10$] minutes, and it is at 50 degrees at #math.equation(block: false, alt: "t equals 20")[$t = 20$] minutes. Compute the ambient temperature.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals 2 x y")[$y^(′) = 2 x y$].


      Answer


      #math.equation(block: true, alt: "y equals C e to the power x squared")[$y = C e^(x^(2))$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x prime equals 3 x t squared minus 3 t squared")[$x^(′) = 3 x t^(2) − 3 t^(2)$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 2")[$x ( 0 ) = 2$].


      Answer


      #math.equation(block: true, alt: "y equals e to the power t cubed plus 1")[$y = e^(t^(3)) + 1$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an implicit solution for #math.equation(block: false, alt: "x prime equals the fraction 1 over 3 x squared plus 1")[$x^(′) = frac(1, 3 x^(2) + 1)$] #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1")[$x ( 0 ) = 1$].


      Answer


      #math.equation(block: true, alt: "x cubed plus x equals t plus 2")[$x^(3) + x = t + 2$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an explicit solution to #math.equation(block: false, alt: "x y prime equals y squared")[$x y^(′) = y^(2)$], #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 1")[$y ( 1 ) = 1$].


      Answer


      #math.equation(block: true, alt: "y equals the fraction 1 over 1 minus ln x")[$y = frac(1, 1 − ln x)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find an implicit solution to #math.equation(block: false, alt: "y prime equals the fraction sin open parenthesis x close parenthesis over cos open parenthesis y close parenthesis")[$y^(′) = frac(sin ( x ), cos ( y ))$].


      Answer


      #math.equation(block: true, alt: "sin open parenthesis y close parenthesis equals minus cos open parenthesis x close parenthesis plus C")[$sin ( y ) = − cos ( x ) + C$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take Example 1.3.3with the same numbers: #math.equation(block: false, alt: "89")[$89$] degrees at #math.equation(block: false, alt: "t equals 0")[$t = 0$], #math.equation(block: false, alt: "85")[$85$] degrees at #math.equation(block: false, alt: "t equals 1")[$t = 1$], and ambient temperature of #math.equation(block: false, alt: "22")[$22$] degrees. Suppose these temperatures were measured with precision of #math.equation(block: false, alt: "± 0.5")[$± 0.5$] degrees. Given this imprecision, the time it takes the coffee to cool to (exactly) 60 degrees is also only known in a certain range. Find this range. Hint: Think about what kind of error makes the cooling time longer and what shorter.


      Answer


      The range is approximately #math.equation(block: false, alt: "7.45")[$7.45$] to #math.equation(block: false, alt: "12.15")[$12.15$] minutes.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A population #math.equation(block: false, alt: "x")[$x$] of rabbits on an island is modeled by #math.equation(block: false, alt: "x prime equals x minus open parenthesis the fraction 1 over 1000 close parenthesis x squared")[$x^(′) = x − ( frac(1, 1000) ) x^(2)$], where the independent variable is time in months. At time #math.equation(block: false, alt: "t equals 0")[$t = 0$], there are #math.equation(block: false, alt: "40")[$40$] rabbits on the island.


      + Find the solution to the equation with the initial condition.
+ How many rabbits are on the island in #math.equation(block: false, alt: "1")[$1$] month, #math.equation(block: false, alt: "5")[$5$] months, #math.equation(block: false, alt: "10")[$10$] months, #math.equation(block: false, alt: "15")[$15$] months (round to the nearest integer)


      Answer


      + #math.equation(block: false, alt: "x equals the fraction 1000 e to the power t over e to the power t plus 24")[$x = frac(1000 e^(t), e^(t) + 24)$].
+ #math.equation(block: false, alt: "102")[$102$] rabbits after one month, #math.equation(block: false, alt: "861")[$861$] after #math.equation(block: false, alt: "5")[$5$] months, #math.equation(block: false, alt: "999")[$999$] after #math.equation(block: false, alt: "10")[$10$] months, #math.equation(block: false, alt: "1000")[$1000$] after #math.equation(block: false, alt: "15")[$15$] months.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.4%3A_Linear_equations_and_the_integrating_factor")[1.4: Linear equations and the integrating factor]
    


    In the exercises, feel free to leave answer as a definite integral if a closed form solution cannot be found. If you can find a closed form solution, you should give that.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus x y equals x")[$y^(′) + x y = x$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus 6 y equals e to the power x")[$y^(′) + 6 y = e^(x)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus 3 x squared y equals sin open parenthesis x close parenthesis e to the power minus x cubed")[$y^(′) + 3 x^(2) y = sin ( x ) e^(− x^(3))$] with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus cos open parenthesis x close parenthesis y equals cos open parenthesis x close parenthesis")[$y^(′) + cos ( x ) y = cos ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "the fraction 1 over x squared plus 1 y prime plus x y equals 3")[$frac(1, x^(2) + 1) y^(′) + x y = 3$] with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 0")[$y ( 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows into the second lake, and then water from the second lake flows further downstream. The in and out flow from each lake is #math.equation(block: false, alt: "500")[$500$] liters per hour. The first lake contains #math.equation(block: false, alt: "100")[$100$] thousand liters of water and the second lake contains #math.equation(block: false, alt: "200")[$200$] thousand liters of water. A truck with #math.equation(block: false, alt: "500")[$500$] kg of toxic substance crashes into the first lake. Assume that the water is being continually mixed perfectly by the stream.


      + Find the concentration of toxic substance as a function of time in both lakes.
+ When will the concentration in the first lake be below #math.equation(block: false, alt: "0.001 kg")[$0.001 " kg"$] per liter?
+ When will the concentration in the second lake be maximal?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Newton’s law of cooling states that #math.equation(block: false, alt: "the fraction d x over d t equals minus k open parenthesis x minus A close parenthesis")[$frac(d x, d t) = − k ( x − A )$]  where #math.equation(block: false, alt: "x")[$x$] is the temperature, #math.equation(block: false, alt: "t")[$t$] is time, #math.equation(block: false, alt: "A")[$A$] is the ambient temperature, and #math.equation(block: false, alt: "k greater than 0")[$k > 0$] is a constant. Suppose that #math.equation(block: false, alt: "A equals A sub 0 cos open parenthesis ω t close parenthesis")[$A = A_(0) cos ( ω t )$] for some constants #math.equation(block: false, alt: "A sub 0")[$A_(0)$] and #math.equation(block: false, alt: "ω")[$ω$]. That is, the ambient temperature oscillates (for example night and day temperatures).


      + Find the general solution.
+ In the long term, will the initial conditions make much of a difference? Why or why not?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Initially #math.equation(block: false, alt: "5")[$5$] grams of salt are dissolved in #math.equation(block: false, alt: "20")[$20$] liters of water. Brine with concentration of salt #math.equation(block: false, alt: "2")[$2$] grams of salt per liter is added at a rate of #math.equation(block: false, alt: "3")[$3$] liters a minute. The tank is mixed well and is drained at #math.equation(block: false, alt: "3")[$3$] liters a minute. How long does the process have to continue until there are #math.equation(block: false, alt: "20")[$20$] grams of salt in the tank?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Initially a tank contains #math.equation(block: false, alt: "10")[$10$] liters of pure water. Brine of unknown (but constant) concentration of salt is flowing in at #math.equation(block: false, alt: "1")[$1$] liter per minute. The water is mixed well and drained at #math.equation(block: false, alt: "1")[$1$] liter per minute. In #math.equation(block: false, alt: "20")[$20$] minutes there are #math.equation(block: false, alt: "15")[$15$] grams of salt in the tank. What is the concentration of salt in the incoming brine?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus 3 x squared y plus x squared")[$y^(′) + 3 x^(2) y + x^(2)$].


      Answer


      #math.equation(block: true, alt: "y equals C e to the power minus x cubed plus the fraction 1 over 3")[$y = C e^(− x^(3)) + frac(1, 3)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus 2 sin open parenthesis 2 x close parenthesis y equals 2 sin open parenthesis 2 x close parenthesis")[$y^(′) + 2 sin ( 2 x ) y = 2 sin ( 2 x )$] with #math.equation(block: false, alt: "y open parenthesis π / 2 close parenthesis equals 3")[$y ( π / 2 ) = 3$].


      Answer


      #math.equation(block: true, alt: "y equals 2 e to the power cos open parenthesis 2 x close parenthesis plus 1 plus 1")[$y = 2 e^(cos ( 2 x ) + 1) + 1$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose a water tank is being pumped out at #math.equation(block: false, alt: "3 the fraction L over min")[$3 #h(0.222em) frac("L", "min")$]. The water tank starts at #math.equation(block: false, alt: "10 L")[$10 " L"$] of clean water. Water with toxic substance is flowing into the tank at #math.equation(block: false, alt: "2 the fraction L over min")[$2 #h(0.222em) frac("L", "min")$], with concentration #math.equation(block: false, alt: "20 t the fraction g over L")[$20 t #h(0.222em) frac("g", "L")$] at time #math.equation(block: false, alt: "t")[$t$]. When the tank is half empty, how many grams of toxic substance are in the tank (assuming perfect mixing)?


      Answer


      #math.equation(block: false, alt: "250")[$250$] grams
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose we have bacteria on a plate and suppose that we are slowly adding a toxic substance such that the rate of growth is slowing down. That is, suppose that #math.equation(block: false, alt: "the fraction d P over d t equals open parenthesis 2 minus 0.1 t close parenthesis P")[$frac(d P, d t) = ( 2 − 0.1 t ) P$]. If #math.equation(block: false, alt: "P open parenthesis 0 close parenthesis equals 1000")[$P ( 0 ) = 1000$], find the population at #math.equation(block: false, alt: "t equals 5")[$t = 5$].


      Answer


      #math.equation(block: true, alt: "P open parenthesis 5 close parenthesis equals 1000 e to the power 2 times 5 minus 0.05 times 5 squared equals 1000 e to the power 8.75 approximately equals 6.31 times 10 to the power 6")[$P ( 5 ) = 1000 e^(2 × 5 − 0.05 × 5^(2)) = 1000 e^(8.75) ≈ 6.31 × 10^(6)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A cylindrical water tank has water flowing in at #math.equation(block: false, alt: "I")[$I$] cubic meters per second. Let #math.equation(block: false, alt: "A")[$A$] be the area of the cross section of the tank in meters. Suppose water is flowing from the bottom of the tank at a rate proportional to the height of the water level. Set up the differential equation for #math.equation(block: false, alt: "h")[$h$], the height of the water, introducing and naming constants that you need. You should also give the units for your constants.


      Answer


      #math.equation(block: false, alt: "A h prime equals I minus k h")[$A h^(′) = I − k h$], where #math.equation(block: false, alt: "k")[$k$] is a constant with units #math.equation(block: false, alt: "the fraction m squared over s")[$frac("m"^(2), "s")$].
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.5%3A_Substitution")[1.5: Substitution]
    


    Hint: Answers need not always be in closed form.


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus y open parenthesis x squared minus 1 close parenthesis plus x y to the power 6 equals 0")[$y^(′) + y ( x^(2) − 1 ) + x y^(6) = 0$], with #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 1")[$y ( 1 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "2 y y prime plus 1 equals y squared plus x")[$2 y y^(′) + 1 = y^(2) + x$], with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime plus x y equals y to the power 4")[$y^(′) + x y = y^(4)$], with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y y prime plus x equals the square root of x squared plus y squared")[$y y^(′) + x = sqrt(x^(2) + y^(2))$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals open parenthesis x plus y minus 1 close parenthesis squared")[$y^(′) = attach(( x + y − 1 ), t: 2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y prime equals the fraction x squared minus y squared over x y")[$y^(′) = frac(x^(2) − y^(2), x y)$], with #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 2")[$y ( 1 ) = 2$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x y prime plus y plus y squared equals 0")[$x y^(′) + y + y^(2) = 0$], #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 2")[$y ( 1 ) = 2$].


      Answer


      #math.equation(block: true, alt: "y equals the fraction 2 over 3 x minus 2")[$y = frac(2, 3 x − 2)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x y prime plus y plus x equals 0")[$x y^(′) + y + x = 0$], #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis equals 1")[$y ( 1 ) = 1$].


      Answer


      #math.equation(block: true, alt: "y equals the fraction 3 minus x squared over 2 x")[$y = frac(3 − x^(2), 2 x)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "y squared y prime equals y cubed minus 3 x")[$y^(2) y^(′) = y^(3) − 3 x$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 2")[$y ( 0 ) = 2$].


      Answer


      #math.equation(block: true, alt: "y equals open parenthesis 7 e to the power 3 x plus 3 x plus 1 close parenthesis to the power 1 / 3")[$y = attach(( 7 e^(3 x) + 3 x + 1 ), t: 1 / 3)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "2 y y prime equals e to the power y squared minus x squared plus 2 x")[$2 y y^(′) = e^(y^(2) − x^(2)) + 2 x$].


      Answer


      #math.equation(block: true, alt: "y equals the square root of x squared minus ln open parenthesis C minus x close parenthesis")[$y = sqrt(x^(2) − ln ( C − x ))$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.6%3A_Autonomous_equations")[1.6: Autonomous equations]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider #math.equation(block: false, alt: "x prime equals x squared")[$x^(′) = x^(2)$].


      + Draw the phase diagram, find the critical points and mark them stable or unstable.
+ Sketch typical solutions of the equation.
+ Find #math.equation(block: false, alt: "lim t → infinity x open parenthesis t close parenthesis")[$limits(lim)_(t → ∞) x ( t )$] for the solution with the initial condition #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals minus 1")[$x ( 0 ) = − 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "x prime equals sin x")[$x^(′) = sin x$].


      + Draw the phase diagram for #math.equation(block: false, alt: "minus 4 π less than or equal to x less than or equal to 4 π")[$− 4 π ≤ x ≤ 4 π$]. On this interval mark the critical points stable or unstable.
+ Sketch typical solutions of the equation.
+ Find #math.equation(block: false, alt: "lim t → infinity x open parenthesis t close parenthesis")[$limits(lim)_(t → ∞) x ( t )$] for the solution with the initial condition #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1")[$x ( 0 ) = 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "f open parenthesis x close parenthesis")[$f ( x )$] is positive for #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$], it is zero when #math.equation(block: false, alt: "x equals 0")[$x = 0$] and #math.equation(block: false, alt: "x equals 1")[$x = 1$], and it is negative for all other #math.equation(block: false, alt: "x")[$x$].


      + Draw the phase diagram for #math.equation(block: false, alt: "x prime equals f open parenthesis x close parenthesis")[$x^(′) = f ( x )$], find the critical points and mark them stable or unstable.
+ Sketch typical solutions of the equation.
+ Find #math.equation(block: false, alt: "lim t → infinity x open parenthesis t close parenthesis")[$limits(lim)_(t → ∞) x ( t )$] for the solution with the initial condition #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0.5")[$x ( 0 ) = 0.5$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Start with the logistic equation #math.equation(block: false, alt: "the fraction d x over d t equals k x open parenthesis M minus x close parenthesis")[$frac(d x, d t) = k x ( M − x )$]. Suppose that we modify our harvesting. That is we will only harvest an amount proportional to current population. In other words we harvest #math.equation(block: false, alt: "h x")[$h x$] per unit of time for some #math.equation(block: false, alt: "h greater than 0")[$h > 0$] (Similar to earlier example with #math.equation(block: false, alt: "h")[$h$] replaced with #math.equation(block: false, alt: "h x")[$h x$]).


      + Construct the differential equation.
+ Show that if #math.equation(block: false, alt: "k M greater than h")[$k M > h$], then the equation is still logistic.
+ What happens when #math.equation(block: false, alt: "k M less than h")[$k M < h$]?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A disease is spreading through the country. Let #math.equation(block: false, alt: "x")[$x$] be the number of people infected. Let the constant #math.equation(block: false, alt: "S")[$S$] be the number of people susceptible to infection. The infection rate #math.equation(block: false, alt: "the fraction d x over d t")[$frac(d x, d t)$] is proportional to the product of already infected people, #math.equation(block: false, alt: "x")[$x$], and the number of susceptible but uninfected people, #math.equation(block: false, alt: "S minus x")[$S − x$].


      + Write down the differential equation.
+ Supposing #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis greater than 0")[$x ( 0 ) > 0$], that is, some people are infected at time #math.equation(block: false, alt: "t equals 0")[$t = 0$], what is #math.equation(block: false, alt: "lim sub t → infinity x open parenthesis t close parenthesis")[$lim_(t → ∞) x ( t )$].
+ Does the solution to part b) agree with your intuition? Why or why not?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "x prime equals open parenthesis x minus 1 close parenthesis open parenthesis x minus 2 close parenthesis x squared")[$x^(′) = ( x − 1 ) ( x − 2 ) x^(2)$].


      + Sketch the phase diagram and find critical points.
+ Classify the critical points.
+ If #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0.5")[$x ( 0 ) = 0.5$] then find #math.equation(block: false, alt: "lim t → infinity x open parenthesis t close parenthesis")[$limits(lim)_(t → ∞) x ( t )$].


      Answer


      + #math.equation(block: false, alt: "0 , 1 , 2")[$0 , #h(0.222em) 1 , #h(0.222em) 2$] are critical points.
+ #math.equation(block: false, alt: "x equals 0")[$x = 0$] is unstable (semistable), #math.equation(block: false, alt: "x equals 1")[$x = 1$] is stable, and #math.equation(block: false, alt: "x equals 2")[$x = 2$] is unstable.
+ #math.equation(block: false, alt: "1")[$1$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "x prime equals e to the power minus x")[$x^(′) = e^(− x)$].


      + Find and classify all critical points.
+ Find #math.equation(block: false, alt: "lim t → infinity x open parenthesis t close parenthesis")[$limits(lim)_(t → ∞) x ( t )$] given any initial condition.


      Answer


      + There are no critical points.
+ #math.equation(block: false, alt: "infinity")[$∞$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Assume that a population of fish in a lake satisfies#math.equation(block: false, alt: "the fraction d x over d t equals k x open parenthesis M minus x close parenthesis")[$frac(d x, d t) = k x ( M − x )$]. Now suppose that fish are continually added at #math.equation(block: false, alt: "A")[$A$] fish per unit of time.


      + Find the differential equation for #math.equation(block: false, alt: "x")[$x$].
+ What is the new limiting population?


      Answer


      + #math.equation(block: false, alt: "the fraction d x over d t equals k x open parenthesis M minus x close parenthesis plus A")[$frac(d x, d t) = k x ( M − x ) + A$]
+ #math.equation(block: false, alt: "the fraction k M plus the square root of open parenthesis k M close parenthesis squared plus 4 A k over 2 k")[$frac(k M + sqrt(( k M )^(2) + 4 A k), 2 k)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose #math.equation(block: false, alt: "the fraction d x over d t equals open parenthesis x minus α close parenthesis open parenthesis x minus β close parenthesis")[$frac(d x, d t) = ( x − α ) ( x − β )$] for two numbers #math.equation(block: false, alt: "α less than β")[$α < β$].


      + Find the critical points, and classify them.


      For b), c), d), find #math.equation(block: false, alt: "lim sub t → infinity x open parenthesis t close parenthesis")[$lim_(t → ∞) x ( t )$] based on the phase diagram.


      + #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis less than α")[$x ( 0 ) < α$],
+ #math.equation(block: false, alt: "α less than x open parenthesis 0 close parenthesis less than β")[$α < x ( 0 ) < β$],
+ #math.equation(block: false, alt: "β less than x open parenthesis 0 close parenthesis")[$β < x ( 0 )$].


      Answer


      + #math.equation(block: false, alt: "α")[$α$] is a stable critical point, #math.equation(block: false, alt: "β")[$β$] is an unstable one.
+ #math.equation(block: false, alt: "α")[$α$]
+ #math.equation(block: false, alt: "α")[$α$]
+ #math.equation(block: false, alt: "infinity")[$∞$] or DNE.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.7%3A_Numerical_methods%3A_Eulers_method")[1.7: Numerical methods: Euler’s method]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider #math.equation(block: false, alt: "the fraction d x over d t equals open parenthesis 2 t minus x close parenthesis squared")[$frac(d x, d t) = ( 2 t − x )^(2)$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 2")[$x ( 0 ) = 2$]. Use Euler’s method with step size #math.equation(block: false, alt: "h equals 0.5")[$h = 0.5$] to approximate #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis")[$x ( 1 )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider #math.equation(block: false, alt: "the fraction d x over d t equals t minus x")[$frac(d x, d t) = t − x$], #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1")[$x ( 0 ) = 1$].


      + Use Euler’s method with step sizes #math.equation(block: false, alt: "h equals 1 , the fraction 1 over 2 , the fraction 1 over 4 , the fraction 1 over 8")[$h = 1 , frac(1, 2) , frac(1, 4) , frac(1, 8)$] to approximate #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis")[$x ( 1 )$].
+ Solve the equation exactly.
+ Describe what happens to the errors for each #math.equation(block: false, alt: "h")[$h$] you used. That is, find the factor by which the error changed each time you halved the interval.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Approximate the value of #math.equation(block: false, alt: "e")[$e$] by looking at the initial value problem #math.equation(block: false, alt: "y prime equals y")[$y^(′) = y$] with #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$] and approximating #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis")[$y ( 1 )$] using Euler’s method with a step size of #math.equation(block: false, alt: "0.2")[$0.2$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Example of numerical instability: Take #math.equation(block: false, alt: "y prime equals minus 5 y")[$y^(′) = − 5 y$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$]. We know that the solution should decay to zero as #math.equation(block: false, alt: "x")[$x$] grows. Using Euler’s method, start with #math.equation(block: false, alt: "h equals 1")[$h = 1$] and compute #math.equation(block: false, alt: "y sub 1 , y sub 2 , y sub 3 , y sub 4")[$y_(1) , y_(2) , y_(3) , y_(4)$] to try to approximate #math.equation(block: false, alt: "y open parenthesis 4 close parenthesis")[$y ( 4 )$]. What happened? Now halve the interval. Keep halving the interval and approximating #math.equation(block: false, alt: "y open parenthesis 4 close parenthesis")[$y ( 4 )$] until the numbers you are getting start to stabilize (that is, until they start going towards zero). Note: You might want to use a calculator.


      The simplest method used in practice is the Runge-Kutta method. Consider #math.equation(block: false, alt: "the fraction d y over d x equals f open parenthesis x , y close parenthesis")[$frac(d y, d x) = f ( x , y )$], #math.equation(block: false, alt: "y open parenthesis x sub 0 close parenthesis equals y sub 0")[$y ( x_(0) ) = y_(0)$] and a step size #math.equation(block: false, alt: "h")[$h$]. Everything is the same as in Euler’s method, except the computation of #math.equation(block: false, alt: "y sub i plus 1")[$y_(i + 1)$] and #math.equation(block: false, alt: "x sub i plus 1")[$x_(i + 1)$].


      #math.equation(block: true, alt: "k sub 1 equals f open parenthesis x sub i , y sub i close parenthesis ,; k sub 2 equals f open parenthesis x sub i plus the fraction h over 2 , y sub i plus k sub 1 the fraction h over 2 close parenthesis x sub i plus 1 equals x sub i plus h ,; k sub 3 equals f open parenthesis x sub i plus the fraction h over 2 , y sub i plus k sub 2 the fraction h over 2 close parenthesis y sub i plus 1 equals y sub i plus the fraction k sub 1 plus 2 k sub 2 plus 2 k sub 3 plus k sub 4 over 6 h ,; k sub 4 equals f open parenthesis x sub i plus h , y sub i plus k sub 3 h close parenthesis .")[$k_(1) = f ( x_(i) , y_(i) ) , \ k_(2) = f ( x_(i) + frac(h, 2) , y_(i) + k_(1) frac(h, 2) ) x_(i + 1) = x_(i) + h , \ k_(3) = f ( x_(i) + frac(h, 2) , y_(i) + k_(2) frac(h, 2) ) y_(i + 1) = y_(i) + frac(k_(1) + 2 k_(2) + 2 k_(3) + k_(4), 6) h , \ k_(4) = f ( x_(i) + h , y_(i) + k_(3) h ) .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider #math.equation(block: false, alt: "the fraction d y over d x equals y x squared")[$frac(d y, d x) = y x^(2)$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].


      + Use Runge-Kutta (see above) with step sizes #math.equation(block: false, alt: "h equals 1")[$h = 1$] and #math.equation(block: false, alt: "h equals the fraction 1 over 2")[$h = frac(1, 2)$] to approximate #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis")[$y ( 1 )$].
+ Use Euler’s method with #math.equation(block: false, alt: "h equals 1")[$h = 1$] and #math.equation(block: false, alt: "h equals the fraction 1 over 2")[$h = frac(1, 2)$].
+ Solve exactly, find the exact value of #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis")[$y ( 1 )$], and compare.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "x prime equals sin open parenthesis x t close parenthesis")[$x^(′) = sin ( x t )$], and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1")[$x ( 0 ) = 1$]. Approximate #math.equation(block: false, alt: "x open parenthesis 1 close parenthesis")[$x ( 1 )$] using Euler’s method with step sizes #math.equation(block: false, alt: "1")[$1$], #math.equation(block: false, alt: "0.5")[$0.5$], #math.equation(block: false, alt: "0.25")[$0.25$]. Use a calculator and compute up to #math.equation(block: false, alt: "4")[$4$] decimal digits.


      Answer


      Approximately: #math.equation(block: false, alt: "1.0000 , 1.2397 , 1.382")[$1.0000 , #h(0.222em) 1.2397 , #h(0.222em) 1.382$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "x prime equals 2 t")[$x^(′) = 2 t$], and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$].


      + Approximate #math.equation(block: false, alt: "x open parenthesis 4 close parenthesis")[$x ( 4 )$] using Euler’s method with step sizes #math.equation(block: false, alt: "4")[$4$], #math.equation(block: false, alt: "2")[$2$], and #math.equation(block: false, alt: "1")[$1$].
+ Solve exactly, and compute the errors.
+ Compute the factor by which the errors changed.


      Answer


      + #math.equation(block: false, alt: "0 , 8 , 12")[$0 , #h(0.222em) 8 , #h(0.222em) 12$]
+ #math.equation(block: false, alt: "x open parenthesis 4 close parenthesis equals 16 ,")[$x ( 4 ) = 16 ,$] so errors are: #math.equation(block: false, alt: "16 , 8 , 4")[$16 , #h(0.222em) 8 , #h(0.222em) 4$].
+ Factors are #math.equation(block: false, alt: "0.5 , 0.5 , 0.5")[$0.5 , #h(0.222em) 0.5 , #h(0.222em) 0.5$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "x prime equals x e to the power x t plus 1")[$x^(′) = x e^(x t + 1)$], and #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$].


      + Approximate #math.equation(block: false, alt: "x open parenthesis 4 close parenthesis")[$x ( 4 )$] using Euler’s method with step sizes #math.equation(block: false, alt: "4")[$4$], #math.equation(block: false, alt: "2")[$2$], and #math.equation(block: false, alt: "1")[$1$].
+ Guess an exact solution based on part a) and compute the errors.


      Answer


      + #math.equation(block: false, alt: "0 , 0 , 0")[$0 , #h(0.222em) 0 , #h(0.222em) 0$]
+ #math.equation(block: false, alt: "x equals 0")[$x = 0$] is a solution so errors are: #math.equation(block: false, alt: "0 , 0 , 0")[$0 , #h(0.222em) 0 , #h(0.222em) 0$].
]


    There is a simple way to improve Euler’s method to make it a second order method by doing just one extra step. Consider #math.equation(block: false, alt: "the fraction d y over d x equals f open parenthesis x , y close parenthesis")[$frac(d y, d x) = f ( x , y )$], #math.equation(block: false, alt: "y open parenthesis x sub 0 close parenthesis equals y sub 0")[$y ( x_(0) ) = y_(0)$], and a step size #math.equation(block: false, alt: "h")[$h$]. What we do is to pretend we compute the next step as in Euler, that is, we start with #math.equation(block: false, alt: "open parenthesis x sub i , y sub i close parenthesis")[$( x_(i) , y_(i) )$], we compute a slope #math.equation(block: false, alt: "k sub 1 equals f open parenthesis x sub i , y sub i close parenthesis")[$k_(1) = f ( x_(i) , y_(i) )$], and then look at the point #math.equation(block: false, alt: "open parenthesis x sub i plus h , y sub i plus k sub 1 h close parenthesis")[$( x_(i) + h , y_(i) + k_(1) h )$]. Instead of letting our new point be #math.equation(block: false, alt: "open parenthesis x sub i plus h , y sub i plus k sub 1 h close parenthesis")[$( x_(i) + h , y_(i) + k_(1) h )$], we compute the slope at that point, call it #math.equation(block: false, alt: "k sub 2")[$k_(2)$], and then take the average of #math.equation(block: false, alt: "k sub 1")[$k_(1)$] and #math.equation(block: false, alt: "k sub 2")[$k_(2)$], hoping that the average is going to be closer to the actual slope on the interval from #math.equation(block: false, alt: "x sub i")[$x_(i)$] to #math.equation(block: false, alt: "x sub i plus h")[$x_(i) + h$]. And we are correct, if we halve the step, the error should go down by a factor of #math.equation(block: false, alt: "2 squared equals 4")[$2^(2) = 4$]. To summarize, the setup is the same as for regular Euler, except the computation of #math.equation(block: false, alt: "y sub i plus 1")[$y_(i + 1)$] and #math.equation(block: false, alt: "x sub i plus 1")[$x_(i + 1)$].


    #math.equation(block: true, alt: "k sub 1 equals f open parenthesis x sub i , y sub i close parenthesis , x sub i plus 1 equals x sub i plus h ,; k sub 2 equals f open parenthesis x sub i plus h , y sub i plus k sub 1 h close parenthesis , y sub i plus 1 equals y sub i plus the fraction k sub 1 plus k sub 2 over 2 h .")[$k_(1) = f ( x_(i) , y_(i) ) , x_(i + 1) = x_(i) + h , \ k_(2) = f ( x_(i) + h , y_(i) + k_(1) h ) , y_(i + 1) = y_(i) + frac(k_(1) + k_(2), 2) h .$]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Consider #math.equation(block: false, alt: "the fraction d y over d x equals x plus y")[$frac(d y, d x) = x + y$], #math.equation(block: false, alt: "y open parenthesis 0 close parenthesis equals 1")[$y ( 0 ) = 1$].


      + Use the improved Euler’s method (see above) with step sizes #math.equation(block: false, alt: "h equals the fraction 1 over 4")[$h = frac(1, 4)$] and #math.equation(block: false, alt: "h equals the fraction 1 over 8")[$h = frac(1, 8)$] to approximate #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis")[$y ( 1 )$].
+ Use Euler’s method with #math.equation(block: false, alt: "h equals the fraction 1 over 4")[$h = frac(1, 4)$] and #math.equation(block: false, alt: "h equals the fraction 1 over 8")[$h = frac(1, 8)$].
+ Solve exactly, find the exact value of #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis")[$y ( 1 )$].
+ Compute the errors, and the factors by which the errors changed.


      Answer


      + Improved Euler: #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis approximately equals 3.3897")[$y ( 1 ) ≈ 3.3897$] for #math.equation(block: false, alt: "h equals 1 / 4")[$h = 1 / 4$], #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis approximately equals 3.4237")[$y ( 1 ) ≈ 3.4237$] for #math.equation(block: false, alt: "h equals 1 / 8")[$h = 1 / 8$],
+ Standard Euler: #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis approximately equals 2.8828")[$y ( 1 ) ≈ 2.8828$] for #math.equation(block: false, alt: "h equals 1 / 4")[$h = 1 / 4$], #math.equation(block: false, alt: "y open parenthesis 1 close parenthesis approximately equals 3.1316")[$y ( 1 ) ≈ 3.1316$] for #math.equation(block: false, alt: "h equals 1 / 8")[$h = 1 / 8$],
+ #math.equation(block: false, alt: "y equals 2 e to the power x minus x minus 1")[$y = 2 e^(x) − x − 1$], so #math.equation(block: false, alt: "y open parenthesis 2 close parenthesis")[$y ( 2 )$] is approximately #math.equation(block: false, alt: "3.4366")[$3.4366$].
+ Approximate errors for improved Euler: #math.equation(block: false, alt: "0.046852")[$0.046852$] for #math.equation(block: false, alt: "h equals 1 / 4")[$h = 1 / 4$], and #math.equation(block: false, alt: "0.012881")[$0.012881$] for #math.equation(block: false, alt: "h equals 1 / 8")[$h = 1 / 8$]. For standard Euler: #math.equation(block: false, alt: "0.55375")[$0.55375$] for #math.equation(block: false, alt: "h equals 1 / 4")[$h = 1 / 4$], and #math.equation(block: false, alt: "0.30499")[$0.30499$] for #math.equation(block: false, alt: "h equals 1 / 8")[$h = 1 / 8$]. Factor is approximately #math.equation(block: false, alt: "0.27")[$0.27$] for improved Euler, and #math.equation(block: false, alt: "0.55")[$0.55$] for standard Euler.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/1%3A_First_order_ODEs/1.8%3A_Exact_Equations")[1.8 Exact Equations]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the following exact equations, implicit general solutions will suffice:


      + #math.equation(block: false, alt: "open parenthesis 2 x y plus x squared close parenthesis d x plus open parenthesis x squared plus y squared plus 1 close parenthesis d y equals 0")[$( 2 x y + x^(2) ) d x + ( x^(2) + y^(2) + 1 ) d y = 0$]
+ #math.equation(block: false, alt: "x to the power 5 plus y to the power 5 the fraction d y over d x equals 0")[$x^(5) + y^(5) frac(d y, d x) = 0$]
+ #math.equation(block: false, alt: "e to the power x plus y cubed plus 3 x y squared the fraction d y over d x equals 0")[$e^(x) + y^(3) + 3 x y^(2) frac(d y, d x) = 0$]
+ #math.equation(block: false, alt: "open parenthesis x plus y close parenthesis cos open parenthesis x close parenthesis plus sin open parenthesis x close parenthesis plus sin open parenthesis x close parenthesis y prime equals 0")[$( x + y ) cos ( x ) + sin ( x ) + sin ( x ) y^(′) = 0$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the integrating factor for the following equations making them into exact equations:


      + #math.equation(block: false, alt: "e to the power x y d x plus the fraction y over x e to the power x y d y equals 0")[$e^(x y) d x + frac(y, x) e^(x y) d y = 0$]
+ #math.equation(block: false, alt: "the fraction e to the power x plus y cubed over y squared d x plus 3 x d y equals 0")[$frac(e^(x) + y^(3), y^(2)) d x + 3 x d y = 0$]
+ #math.equation(block: false, alt: "4 open parenthesis y squared plus x close parenthesis d x plus the fraction 2 x plus 2 y squared over y d y equals 0")[$4 ( y^(2) + x ) d x + frac(2 x + 2 y^(2), y) d y = 0$]
+ #math.equation(block: false, alt: "2 sin open parenthesis y close parenthesis d x plus x cos open parenthesis y close parenthesis d y equals 0")[$2 sin ( y ) d x + x cos ( y ) d y = 0$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you have an equation of the form: #math.equation(block: false, alt: "f open parenthesis x close parenthesis plus g open parenthesis y close parenthesis the fraction d y over d x equals 0")[$f ( x ) + g ( y ) frac(d y, d x) = 0$].


      + Show it is exact.
+ Find the form of the potential function in terms of #math.equation(block: false, alt: "f")[$f$] and #math.equation(block: false, alt: "g")[$g$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that we have the equation #math.equation(block: false, alt: "f open parenthesis x close parenthesis d x minus d y equals 0")[$f ( x ) d x − d y = 0$].


      + Is this equation exact?
+ Find the general solution using a definite integral.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the potential function #math.equation(block: false, alt: "F open parenthesis x , y close parenthesis")[$F ( x , y )$] of the exact equation #math.equation(block: false, alt: "the fraction 1 plus x y over x d x plus open parenthesis the fraction 1 over y plus x close parenthesis d y equals 0")[$frac(1 + x y, x) d x + ( frac(1, y) + x ) d y = 0$] in two different ways.


      + Integrate #math.equation(block: false, alt: "M")[$M$] in terms of #math.equation(block: false, alt: "x")[$x$] and then differentiate in #math.equation(block: false, alt: "y")[$y$] and set to #math.equation(block: false, alt: "N")[$N$].
+ Integrate #math.equation(block: false, alt: "N")[$N$] in terms of #math.equation(block: false, alt: "y")[$y$] and then differentiate in #math.equation(block: false, alt: "x")[$x$] and set to #math.equation(block: false, alt: "M")[$M$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
A function #math.equation(block: false, alt: "u open parenthesis x , y close parenthesis")[$u ( x , y )$] is said to be a if #math.equation(block: false, alt: "u sub x x plus u sub y y equals 0")[$u_(x x) + u_(y y) = 0$].


      + Show if #math.equation(block: false, alt: "u")[$u$] is harmonic, #math.equation(block: false, alt: "minus u sub y d x plus u sub x d y equals 0")[$− u_(y) d x + u_(x) d y = 0$] is an exact equation. So there exists (at least locally) the so-called function #math.equation(block: false, alt: "v open parenthesis x , y close parenthesis")[$v ( x , y )$] such that #math.equation(block: false, alt: "v sub x equals minus u sub y")[$v_(x) = − u_(y)$] and #math.equation(block: false, alt: "v sub y equals u sub x")[$v_(y) = u_(x)$].


      Verify that the following #math.equation(block: false, alt: "u")[$u$] are harmonic and find the corresponding harmonic conjugates #math.equation(block: false, alt: "v")[$v$]:


      + #math.equation(block: false, alt: "u equals 2 x y")[$u = 2 x y$]
+ #math.equation(block: false, alt: "u equals e to the power x cos y")[$u = e^(x) cos y$]
+ #math.equation(block: false, alt: "u equals x cubed minus 3 x y squared")[$u = x^(3) − 3 x y^(2)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve the following exact equations, implicit general solutions will suffice:


      + #math.equation(block: false, alt: "cos open parenthesis x close parenthesis plus y e to the power x y plus x e to the power x y y prime equals 0")[$cos ( x ) + y e^(x y) + x e^(x y) y^(′) = 0$]
+ #math.equation(block: false, alt: "open parenthesis 2 x plus y close parenthesis d x plus open parenthesis x minus 4 y close parenthesis d y equals 0")[$( 2 x + y ) d x + ( x − 4 y ) d y = 0$]
+ #math.equation(block: false, alt: "e to the power x plus e to the power y the fraction d y over d x equals 0")[$e^(x) + e^(y) frac(d y, d x) = 0$]
+ #math.equation(block: false, alt: "open parenthesis 3 x squared plus 3 y close parenthesis d x plus open parenthesis 3 y squared plus 3 x close parenthesis d y equals 0")[$( 3 x^(2) + 3 y ) d x + ( 3 y^(2) + 3 x ) d y = 0$]


      Answer


      + #math.equation(block: false, alt: "e to the power x y plus sin open parenthesis x close parenthesis equals C")[$e^(x y) + sin ( x ) = C$]
+ #math.equation(block: false, alt: "x squared plus x y minus 2 y squared equals C")[$x^(2) + x y − 2 y^(2) = C$]
+ #math.equation(block: false, alt: "e to the power x plus e to the power y equals C")[$e^(x) + e^(y) = C$]
+ #math.equation(block: false, alt: "x cubed plus 3 x y plus y cubed equals C")[$x^(3) + 3 x y + y^(3) = C$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the integrating factor for the following equations making them into exact equations:


      + #math.equation(block: false, alt: "the fraction 1 over y d x plus 3 y d y equals 0")[$frac(1, y) d x + 3 y d y = 0$]
+ #math.equation(block: false, alt: "d x minus e to the power minus x minus y d y equals 0")[$d x − e^(− x − y) d y = 0$]
+ #math.equation(block: false, alt: "open parenthesis the fraction cos open parenthesis x close parenthesis over y squared plus the fraction 1 over y close parenthesis d x plus the fraction x over y squared d y equals 0")[$( frac(cos ( x ), y^(2)) + frac(1, y) ) d x + frac(x, y^(2)) d y = 0$]
+ #math.equation(block: false, alt: "open parenthesis 2 y plus the fraction y squared over x close parenthesis d x plus open parenthesis 2 y plus x close parenthesis d y equals 0")[$( 2 y + frac(y^(2), x) ) d x + ( 2 y + x ) d y = 0$]


      Answer


      + Integrating factor is #math.equation(block: false, alt: "y")[$y$], equation becomes #math.equation(block: false, alt: "d x plus 3 y squared d y equals 0")[$d x + 3 y^(2) d y = 0$].
+ Integrating factor is #math.equation(block: false, alt: "e to the power x")[$e^(x)$], equation becomes #math.equation(block: false, alt: "e to the power x d x minus e to the power minus y d y equals 0")[$e^(x) d x − e^(− y) d y = 0$].
+ Integrating factor is #math.equation(block: false, alt: "y squared")[$y^(2)$], equation becomes #math.equation(block: false, alt: "open parenthesis cos open parenthesis x close parenthesis plus y close parenthesis d x plus x d y equals 0")[$( cos ( x ) + y ) d x + x d y = 0$].
+ Integrating factor is #math.equation(block: false, alt: "x")[$x$], equation becomes #math.equation(block: false, alt: "open parenthesis 2 x y plus y squared close parenthesis d x plus open parenthesis x squared plus 2 x y close parenthesis d y equals 0")[$( 2 x y + y^(2) ) d x + ( x^(2) + 2 x y ) d y = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ Show that every separable equation #math.equation(block: false, alt: "y prime equals f open parenthesis x close parenthesis g open parenthesis y close parenthesis")[$y^(′) = f ( x ) g ( y )$] can be written as an exact equation, and verify that it is indeed exact.
+ Using this rewrite #math.equation(block: false, alt: "y prime equals x y")[$y^(′) = x y$] as an exact equation, solve it and verify that the solution is the same as it was in Example 1.3.1.


      Answer


      + The equation is #math.equation(block: false, alt: "minus f open parenthesis x close parenthesis d x plus the fraction 1 over g open parenthesis y close parenthesis d y")[$− f ( x ) d x + frac(1, g ( y )) d y$], and this is exact because #math.equation(block: false, alt: "M equals minus f open parenthesis x close parenthesis")[$M = − f ( x )$], #math.equation(block: false, alt: "N equals the fraction 1 over g open parenthesis y close parenthesis")[$N = frac(1, g ( y ))$], so #math.equation(block: false, alt: "M sub y equals 0 equals N sub x")[$M_(y) = 0 = N_(x)$].
+ #math.equation(block: false, alt: "minus x d x plus the fraction 1 over y d y equals 0")[$− x d x + frac(1, y) d y = 0$], leads to potential function #math.equation(block: false, alt: "F open parenthesis x , y close parenthesis equals minus the fraction x squared over 2 plus ln | y |")[$F ( x , y ) = − frac(x^(2), 2) + ln | y |$], solving #math.equation(block: false, alt: "F open parenthesis x , y close parenthesis equals C")[$F ( x , y ) = C$] leads to the same solution as the example.
]


    
=== 1.9: First Order Linear PDE


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      + #math.equation(block: false, alt: "u sub t plus 9 u sub x equals 0")[$u_(t) + 9 u_(x) = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis")[$u ( x , 0 ) = sin ( x )$],
+ #math.equation(block: false, alt: "u sub t minus 8 u sub x equals 0")[$u_(t) − 8 u_(x) = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis")[$u ( x , 0 ) = sin ( x )$],
+ #math.equation(block: false, alt: "u sub t plus π u sub x equals 0")[$u_(t) + π u_(x) = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis")[$u ( x , 0 ) = sin ( x )$],
+ #math.equation(block: false, alt: "u sub t plus π u sub x plus u equals 0")[$u_(t) + π u_(x) + u = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis")[$u ( x , 0 ) = sin ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "u sub t plus 3 u sub x equals 1")[$u_(t) + 3 u_(x) = 1$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals x squared")[$u ( x , 0 ) = x^(2)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "u sub t plus 3 u sub x equals x")[$u_(t) + 3 u_(x) = x$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals e to the power x")[$u ( x , 0 ) = e^(x)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "u sub x plus u sub t plus x u equals 0")[$u_(x) + u_(t) + x u = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals cos open parenthesis x close parenthesis")[$u ( x , 0 ) = cos ( x )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
+ Find the characteristic coordinates for the following equations:  #linebreak() 
    #math.equation(block: false, alt: "u sub x plus u sub t plus u equals 1")[$u_(x) + u_(t) + u = 1$],  #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals cos open parenthesis x close parenthesis")[$u ( x , 0 ) = cos ( x )$], )  #math.equation(block: false, alt: "2 u sub x plus 2 u sub t plus 2 u equals 2")[$2 u_(x) + 2 u_(t) + 2 u = 2$],  #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals cos open parenthesis x close parenthesis")[$u ( x , 0 ) = cos ( x )$].
+ Solve the two equations using the coordinates.
+ Explain why you got the same solution, although the characteristic coordinates you found were different.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "open parenthesis 1 plus x squared close parenthesis u sub t plus x squared u sub x plus e to the power x u equals 0")[$( 1 + x^(2) ) u_(t) + x^(2) u_(x) + e^(x) u = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals 0")[$u ( x , 0 ) = 0$]. Hint: Think a little out of the box.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve


      + #math.equation(block: false, alt: "u sub t minus 5 u sub x equals 0")[$u_(t) − 5 u_(x) = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals the fraction 1 over 1 plus x squared")[$u ( x , 0 ) = frac(1, 1 + x^(2))$],
+ #math.equation(block: false, alt: "u sub t plus 2 u sub x equals 0")[$u_(t) + 2 u_(x) = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals cos open parenthesis x close parenthesis")[$u ( x , 0 ) = cos ( x )$].


      Answer


      + #math.equation(block: false, alt: "u equals the fraction 1 over 1 plus open parenthesis x plus 5 t close parenthesis squared")[$u = frac(1, 1 + ( x + 5 t )^(2))$]
+ #math.equation(block: false, alt: "u equals cos open parenthesis x minus 2 t close parenthesis")[$u = cos ( x − 2 t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "u sub x plus u sub t plus t u equals 0")[$u_(x) + u_(t) + t u = 0$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals cos open parenthesis x close parenthesis")[$u ( x , 0 ) = cos ( x )$].


      Answer


      #math.equation(block: true, alt: "u equals cos open parenthesis x minus t close parenthesis e to the power minus t squared / 2")[$u = cos ( x − t ) e^(− t^(2) / 2)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "u sub x plus u sub t equals 5")[$u_(x) + u_(t) = 5$], #math.equation(block: false, alt: "u open parenthesis x , 0 close parenthesis equals x")[$u ( x , 0 ) = x$].


      Answer


      #math.equation(block: true, alt: "u equals x plus 4 t")[$u = x + 4 t$]
]