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== 5.4#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)/5%3A_Eigenvalue_problems/5.1%3A_Sturm-Liouville_problems")[5.1: Sturm-Liouville problems]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find eigenvalues and eigenfunctions of


      #math.equation(block: true, alt: "y double prime plus λ y equals 0 , y open parenthesis 0 close parenthesis minus y prime open parenthesis 0 close parenthesis equals 0 , y open parenthesis 1 close parenthesis equals 0 .")[$y^(″) + λ y = 0 , " " " " " " y ( 0 ) − y^(′) ( 0 ) = 0 , " " " " " " y ( 1 ) = 0 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Expand the function #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x")[$f ( x ) = x$] on #math.equation(block: false, alt: "0 less than or equal to x less than or equal to 1")[$0 ≤ x ≤ 1$] using the eigenfunctions of the system


      #math.equation(block: true, alt: "y double prime plus λ y equals 0 , y prime open parenthesis 0 close parenthesis equals 0 , y open parenthesis 1 close parenthesis equals 0 .")[$y^(″) + λ y = 0 , " " " " " " y^(′) ( 0 ) = 0 , " " " " " " y ( 1 ) = 0 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that you had a Sturm-Liouville problem on the interval #math.equation(block: false, alt: "[ 0 , 1 ]")[$[ 0 , 1 ]$] and came up with #math.equation(block: false, alt: "y sub n open parenthesis x close parenthesis equals sin open parenthesis γ n x close parenthesis")[$y_(n) ( x ) = sin ( γ n x )$], where #math.equation(block: false, alt: "γ greater than 0")[$γ > 0$] is some constant. Decompose #math.equation(block: false, alt: "f open parenthesis x close parenthesis equals x , 0 less than x less than 1")[$f ( x ) = x , 0 < x < 1$], in terms of these eigenfunctions.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find eigenvalues and eigenfunctions of


      #math.equation(block: true, alt: "y to the power ′ open parenthesis 4 close parenthesis plus λ y equals 0 , y open parenthesis 0 close parenthesis equals 0 , y prime open parenthesis 0 close parenthesis equals 0 , y open parenthesis 1 close parenthesis equals 0 y prime open parenthesis 1 close parenthesis equals 0 .")[$y^(′ ( 4 )) + λ y = 0 , " " " " " " y ( 0 ) = 0 , " " " " " " y^(′) ( 0 ) = 0 , " " " " " " y ( 1 ) = 0 " " " " " " y^(′) ( 1 ) = 0 .$]


      This problem is not a Sturm-Liouville problem, but the idea is the same.
]


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


      Find eigenvalues and eigenfunctions for


      #math.equation(block: true, alt: "the fraction d over d x open parenthesis e to the power x y prime close parenthesis plus λ e to the power x y equals 0 , y open parenthesis 0 close parenthesis equals 0 , y open parenthesis 1 close parenthesis equals 0 .")[$frac(d, d x) ( e^(x) y^(′) ) + λ e^(x) y = 0 , " " " " " " y ( 0 ) = 0 , " " " " " " y ( 1 ) = 0 .$]


      Hint: First write the system as a constant coefficient system to find general solutions. Do note that Theorem 5.1.1 guarantees #math.equation(block: false, alt: "λ greater than or equal to 0")[$λ ≥ 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find eigenvalues and eigenfunctions of


      #math.equation(block: true, alt: "y double prime plus λ y equals 0 , y open parenthesis minus 1 close parenthesis equals 0 , y open parenthesis 1 close parenthesis equals 0 .")[$y^(″) + λ y = 0 , " " " " " " y ( − 1 ) = 0 , " " " " " " y ( 1 ) = 0 .$]


      Answer


      #math.equation(block: true, alt: "λ sub n equals the fraction open parenthesis 2 n minus 1 close parenthesis π over 2 , n equals 1 , 2 , 3 , ⋯ ,")[$λ_(n) = frac(( 2 n − 1 ) π, 2) , #h(0.222em) n = 1 , #h(0.222em) 2 , #h(0.222em) 3 , ⋯ ,$] #math.equation(block: true, alt: "y sub n equals cos open parenthesis the fraction open parenthesis 2 n minus 1 close parenthesis π over 2 x close parenthesis")[$y_(n) = cos ( frac(( 2 n − 1 ) π, 2) x )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Put the following problems into the standard form for Sturm-Liouville problems, that is, find #math.equation(block: false, alt: "p open parenthesis x close parenthesis , q open parenthesis x close parenthesis , r open parenthesis x close parenthesis , α sub 1 , α sub , β sub 1 , β sub 1 ,")[$p ( x ) , q ( x ) , r ( x ) , α_(1) , α_(,) β_(1) , β_(1) ,$], and decide if the problems are regular or not.


      + #math.equation(block: false, alt: "x y double prime plus λ y equals 0")[$x y^(″) + λ y = 0$] for #math.equation(block: false, alt: "0 less than x less than 1 , y open parenthesis 0 close parenthesis equals 0 , y open parenthesis 1 close parenthesis equals 0 ,")[$0 < x < 1 , #h(0.222em) y ( 0 ) = 0 , #h(0.222em) y ( 1 ) = 0 ,$]
+ #math.equation(block: false, alt: "open parenthesis 1 plus x squared close parenthesis y double prime plus 2 x y prime plus open parenthesis λ minus x squared close parenthesis y equals 0")[$( 1 + x^(2) ) y^(″) + 2 x y^(′) + ( λ − x^(2) ) y = 0$] for #math.equation(block: false, alt: "minus 1 less than x less than 1 , y open parenthesis minus 1 close parenthesis equals 0 , y open parenthesis 1 close parenthesis plus y prime open parenthesis 1 close parenthesis equals 0")[$− 1 < x < 1 , #h(0.222em) y ( − 1 ) = 0 , #h(0.222em) y ( 1 ) + y^(′) ( 1 ) = 0$]


      Answer


      + #math.equation(block: false, alt: "p open parenthesis x close parenthesis equals 1 , q open parenthesis x close parenthesis equals 0 , r open parenthesis x close parenthesis equals the fraction 1 over x , α sub 1 equals 1 , α sub 2 equals 0 , β sub 1 equals 1 , β sub 2 equals 0")[$p ( x ) = 1 , #h(0.222em) q ( x ) = 0 , #h(0.222em) r ( x ) = frac(1, x) , #h(0.222em) α_(1) = 1 , #h(0.222em) α_(2) = 0 , #h(0.222em) β_(1) = 1 , #h(0.222em) β_(2) = 0$]. The problem is not regular.
+ #math.equation(block: false, alt: "p open parenthesis x close parenthesis equals 1 plus x squared , q open parenthesis x close parenthesis equals x squared , r open parenthesis x close parenthesis equals 1 , α sub 1 equals 1 , α sub 2 equals 0 , β sub 1 equals 1 , β sub 2 equals 1")[$p ( x ) = 1 + x^(2) , #h(0.222em) q ( x ) = x^(2) , #h(0.222em) r ( x ) = 1 , #h(0.222em) α_(1) = 1 , #h(0.222em) α_(2) = 0 , #h(0.222em) β_(1) = 1 , #h(0.222em) β_(2) = 1$]. The problem is regular.
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/5%3A_Eigenvalue_problems/5.2%3A_Application_of_Eigenfunction_Series")[5.2: Application of eigenfunction series]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you have a beam of length #math.equation(block: false, alt: "5")[$5$] with free ends. Let #math.equation(block: false, alt: "y")[$y$] be the transverse deviation of the beam at position #math.equation(block: false, alt: "x")[$x$] on the beam #math.equation(block: false, alt: "open parenthesis 0 less than x less than 5 close parenthesis")[$( 0 < x < 5 )$]. You know that the constants are such that this satisfies the equation #math.equation(block: false, alt: "y sub t t plus 4 y sub x x x x equals 0")[$y_(t t) + 4 y_(x x x x) = 0$]. Suppose you know that the initial shape of the beam is the graph of #math.equation(block: false, alt: "x open parenthesis 5 minus x close parenthesis")[$x ( 5 − x )$], and the initial velocity is uniformly equal to #math.equation(block: false, alt: "2")[$2$] (same for each #math.equation(block: false, alt: "x")[$x$]) in the positive #math.equation(block: false, alt: "y")[$y$] direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you have a beam of length #math.equation(block: false, alt: "5")[$5$] with one end free and one end fixed (the fixed end is at #math.equation(block: false, alt: "x equals 5")[$x = 5$]). Let #math.equation(block: false, alt: "u")[$u$] be the longitudinal deviation of the beam at position #math.equation(block: false, alt: "x")[$x$] on the beam #math.equation(block: false, alt: "open parenthesis 0 less than x less than 5 close parenthesis")[$( 0 < x < 5 )$]. You know that the constants are such that this satisfies the equation #math.equation(block: false, alt: "u sub t t equals 4 u sub x x")[$u_(t t) = 4 u_(x x)$]. Suppose you know that the initial displacement of the beam is #math.equation(block: false, alt: "the fraction x minus 5 over 50")[$frac(x − 5, 50)$], and the initial velocity is #math.equation(block: false, alt: "the fraction minus open parenthesis x minus 5 close parenthesis over 100")[$frac(− ( x − 5 ), 100)$] in the positive #math.equation(block: false, alt: "u")[$u$] direction. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose the beam is #math.equation(block: false, alt: "L")[$L$] units long, everything else kept the same as in (5.2.2). What is the equation and the series solution?
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you have


      #math.equation(block: true, alt: "a to the power 4 y sub x x x x plus y sub t t equals 0 open parenthesis 0 less than x less than 1 , t greater than 0 close parenthesis ,; y open parenthesis 0 , t close parenthesis equals y sub x x open parenthesis 0 , t close parenthesis equals 0 ,; y open parenthesis 1 , t close parenthesis equals y sub x x open parenthesis 1 , t close parenthesis equals 0 ,; y open parenthesis x , 0 close parenthesis equals f open parenthesis x close parenthesis , y sub t open parenthesis x , 0 close parenthesis equals g open parenthesis x close parenthesis .")[$a^(4) y_(x x x x) + y_(t t) = 0 #h(1em) ( 0 < x < 1 , t > 0 ) , \ y ( 0 , t ) = y_(x x) ( 0 , t ) = 0 , \ y ( 1 , t ) = y_(x x) ( 1 , t ) = 0 , \ y ( x , 0 ) = f ( x ) , #h(1em) y_(t) ( x , 0 ) = g ( x ) .$]


      That is, you have also an initial velocity. Find a series solution. Hint: Use the same idea as we did for the wave equation.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you have a beam of length #math.equation(block: false, alt: "1")[$1$] with hinged ends. Let #math.equation(block: false, alt: "y")[$y$] be the transverse deviation of the beam at position #math.equation(block: false, alt: "x")[$x$] on the beam (#math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$]). You know that the constants are such that this satisfies the equation #math.equation(block: false, alt: "y sub t t plus 4 y sub x x x x equals 0")[$y_(t t) + 4 y_(x x x x) = 0$]. Suppose you know that the initial shape of the beam is the graph of #math.equation(block: false, alt: "sin open parenthesis π x close parenthesis")[$sin ( π x )$], and the initial velocity is #math.equation(block: false, alt: "0")[$0$]. Solve for #math.equation(block: false, alt: "y")[$y$].


      Answer


      #math.equation(block: true, alt: "y open parenthesis x , t close parenthesis equals sin open parenthesis π x close parenthesis cos open parenthesis 2 π squared t close parenthesis")[$y ( x , t ) = sin ( π x ) cos ( 2 π^(2) t )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose you have a beam of length #math.equation(block: false, alt: "10")[$10$] with two fixed ends. Let #math.equation(block: false, alt: "y")[$y$] be the transverse deviation of the beam at position #math.equation(block: false, alt: "x")[$x$] on the beam (#math.equation(block: false, alt: "0 less than x less than 10")[$0 < x < 10$]). You know that the constants are such that this satisfies the equation #math.equation(block: false, alt: "y sub t t plus 9 y sub x x x x equals 0")[$y_(t t) + 9 y_(x x x x) = 0$]. Suppose you know that the initial shape of the beam is the graph of #math.equation(block: false, alt: "sin open parenthesis π x close parenthesis")[$sin ( π x )$], and the initial velocity is uniformly equal to #math.equation(block: false, alt: "x open parenthesis 10 minus x close parenthesis")[$x ( 10 − x )$]. Set up the equation together with the boundary and initial conditions. Just set up, do not solve.


      Answer


      #math.equation(block: true, alt: "9 y sub x x x x plus y sub t t equals 0")[$9 y_(x x x x) + y_(t t) = 0$] #math.equation(block: true, alt: "open parenthesis 0 less than x less than 10 , t greater than 0 close parenthesis")[$( 0 < x < 10 , #h(0.222em) t > 0 )$], #math.equation(block: true, alt: "y open parenthesis 0 , t close parenthesis equals y sub x open parenthesis 0 , t close parenthesis equals 0")[$#h(1em) y ( 0 , t ) = y_(x) ( 0 , t ) = 0$], #math.equation(block: true, alt: "y open parenthesis 10 , t close parenthesis equals y sub x open parenthesis 10 , t close parenthesis equals 0")[$#h(1em) y ( 10 , t ) = y_(x) ( 10 , t ) = 0$], #math.equation(block: true, alt: "y open parenthesis x , 0 close parenthesis equals sin open parenthesis π x close parenthesis")[$#h(1em) y ( x , 0 ) = sin ( π x )$], #math.equation(block: true, alt: "y sub t open parenthesis x , 0 close parenthesis equals x open parenthesis 10 minus x close parenthesis")[$#h(1em) y_(t) ( x , 0 ) = x ( 10 − x )$].
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/5%3A_Eigenvalue_problems/5.3%3A_Steady_Periodic_Solutions")[5.3: Steady periodic solutions]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Suppose that the forcing function for the vibrating string is #math.equation(block: false, alt: "F sub 0 sin open parenthesis ω t close parenthesis")[$F_(0) sin ( ω t )$]. Derive the particular solution #math.equation(block: false, alt: "y sub p")[$y_(p)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take the forced vibrating string. Suppose that #math.equation(block: false, alt: "L equals 1 , a equals 1")[$L = 1 , a = 1$]. Suppose that the forcing function is the square wave that is #math.equation(block: false, alt: "1")[$1$] on the interval #math.equation(block: false, alt: "0 less than x less than 1")[$0 < x < 1$] and #math.equation(block: false, alt: "minus 1")[$− 1$]on the interval #math.equation(block: false, alt: "minus 1 less than x less than 0")[$− 1 < x < 0$]. Find the particular solution. Hint: You may want to use result of Exercise #math.equation(block: false, alt: "5.3 .1")[$5.3 ".1"$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The units are cgs (centimeters-grams-seconds). For #math.equation(block: false, alt: "k equals 0.005 , ω equals 1.991 times 10 to the power minus 7 , A sub 0 equals 20")[$k = 0.005 , ω = 1.991 × 10^(− 7) , A_(0) = 20$]. Find the depth at which the temperature variation is half (#math.equation(block: false, alt: "± 10")[$± 10$] degrees) of what it is on the surface.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Derive the solution for underground temperature oscillation without assuming that #math.equation(block: false, alt: "T sub 0 equals 0")[$T_(0) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Take the forced vibrating string. Suppose that #math.equation(block: false, alt: "L equals 1 , a equals 1")[$L = 1 , a = 1$]. Suppose that the forcing function is a sawtooth, that is #math.equation(block: false, alt: "| x | minus the fraction 1 over 2")[$| x | − frac(1, 2)$] on #math.equation(block: false, alt: "minus 1 less than x less than 1")[$− 1 < x < 1$] extended periodically. Find the particular solution.


      Answer


      #math.equation(block: true, alt: "y sub p open parenthesis x , t close parenthesis equals ∑ n odd n equals 1 infinity the fraction minus 4 over n to the power 4 π to the power 4 open parenthesis cos open parenthesis n π x close parenthesis minus the fraction cos open parenthesis n π close parenthesis minus 1 over sin open parenthesis n π close parenthesis sin open parenthesis n π x close parenthesis minus 1 close parenthesis cos open parenthesis n π t close parenthesis .")[$y_(p) ( x , t ) = limits(∑)^(∞)_(limits(n " odd")^(n = 1)) frac(− 4, n^(4) π^(4)) ( cos ( n π x ) − frac(cos ( n π ) − 1, sin ( n π )) sin ( n π x ) − 1 ) cos ( n π t ) .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
The units are cgs (centimeters-grams-seconds). For #math.equation(block: false, alt: "k equals 0.01 , ω equals 1.991 times 10 to the power minus 7 , A sub 0 equals 25")[$k = 0.01 , ω = 1.991 × 10^(− 7) , A_(0) = 25$]. Find the depth at which the summer is again the hottest point.


      Answer


      Approximately 1991 centimeters
]