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== 6.6#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)/6%3A_The_Laplace_Transform/6.1%3A_The_Laplace_Transform")[6.1: The Laplace transform]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "3 plus t to the power 5 plus sin open parenthesis π t close parenthesis")[$3 + t^(5) + sin ( π t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "a plus b t plus c t squared")[$a + b t + c t^(2)$] for some constants #math.equation(block: false, alt: "a")[$a$], #math.equation(block: false, alt: "b")[$b$], and #math.equation(block: false, alt: "c")[$c$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "A cos open parenthesis ω t close parenthesis plus B sin open parenthesis ω t close parenthesis")[$A cos ( ω t ) + B sin ( ω t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "cos squared open parenthesis ω t close parenthesis")[$cos^(2) ( ω t )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse Laplace transform of #math.equation(block: false, alt: "the fraction 4 over s squared minus 9")[$frac(4, s^(2) − 9)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse Laplace transform of #math.equation(block: false, alt: "the fraction 2 s over s squared minus 1")[$frac(2 s, s^(2) − 1)$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse Laplace transform of #math.equation(block: false, alt: "the fraction 1 over open parenthesis s minus 1 close parenthesis squared open parenthesis s plus 1 close parenthesis")[$frac(1, ( s − 1 )^(2) ( s + 1 ))$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "t i f t greater than or equal to 1 ,; 0 i f t less than 1 .")[$t upright(i) upright(f) " " t ≥ 1 , \ 0 upright(i) upright(f) " " t < 1 .$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse Laplace transform of #math.equation(block: false, alt: "the fraction s over open parenthesis s squared plus s plus 2 close parenthesis open parenthesis s plus 4 close parenthesis")[$frac(s, ( s^(2) + s + 2 ) ( s + 4 ))$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "sin open parenthesis ω open parenthesis t minus a close parenthesis close parenthesis")[$sin ( ω ( t − a ) )$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "t sin open parenthesis ω t close parenthesis")[$t sin ( ω t )$]. Hint: Several integrations by parts.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "4 open parenthesis t plus 1 close parenthesis squared")[$4 ( t + 1 )^(2)$].


      Answer


      #math.equation(block: true, alt: "the fraction 8 over s cubed plus the fraction 8 over s squared plus the fraction 4 over s")[$frac(8, s^(3)) + frac(8, s^(2)) + frac(4, s)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the inverse Laplace transform of #math.equation(block: false, alt: "the fraction 8 over s cubed open parenthesis s plus 2 close parenthesis")[$frac(8, s^(3) ( s + 2 ))$].


      Answer


      #math.equation(block: true, alt: "2 t squared minus 2 t plus 1 minus e to the power minus 2 t")[$2 t^(2) − 2 t + 1 − e^(− 2 t)$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "t e to the power minus t")[$t e^(− t)$] (Hint: integrate by parts).


      Answer


      #math.equation(block: true, alt: "the fraction 1 over open parenthesis s plus 1 close parenthesis squared")[$frac(1, ( s + 1 )^(2))$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the Laplace transform of #math.equation(block: false, alt: "sin open parenthesis t close parenthesis e to the power minus t")[$sin ( t ) e^(− t)$] (Hint: integrate by parts).


      Answer


      #math.equation(block: true, alt: "the fraction 1 over s squared plus 2 s plus 2")[$frac(1, s^(2) + 2 s + 2)$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/6%3A_The_Laplace_Transform/6.2%3A_Transforms_of_derivatives_and_ODEs")[6.2: Transforms of Derivatives and ODEs]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Verify Table 6.2.1.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the Heaviside function write down the piecewise function that is #math.equation(block: false, alt: "0")[$0$] for #math.equation(block: false, alt: "t less than 0 , t squared")[$t < 0 , t^(2)$] for #math.equation(block: false, alt: "t")[$t$] in #math.equation(block: false, alt: "[ 0 , 1 ]")[$[ 0 , 1 ]$] and #math.equation(block: false, alt: "t")[$t$] for #math.equation(block: false, alt: "t greater than 1")[$t > 1$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the Laplace transform solve


      #math.equation(block: true, alt: "m x double prime plus c x prime plus k x equals 0 , x open parenthesis 0 close parenthesis equals a , x prime open parenthesis 0 close parenthesis equals b .")[$m x^(″) + c x^(′) + k x = 0 , #h(1em) x ( 0 ) = a , #h(1em) x^(′) ( 0 ) = b .$]


      where #math.equation(block: false, alt: "m greater than 0 , c greater than 0 , k greater than 0")[$m > 0 , c > 0 , k > 0$], and #math.equation(block: false, alt: "c squared minus 4 k m greater than 0")[$c^(2) − 4 k m > 0$] (system is overdamped).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the Laplace transform solve


      #math.equation(block: true, alt: "m x double prime plus c x prime plus k x equals 0 , x open parenthesis 0 close parenthesis equals a , x prime open parenthesis 0 close parenthesis equals b .")[$m x^(″) + c x^(′) + k x = 0 , #h(1em) x ( 0 ) = a , #h(1em) x^(′) ( 0 ) = b .$]


      where #math.equation(block: false, alt: "m greater than 0 , c greater than 0 , k greater than 0")[$m > 0 , c > 0 , k > 0$], and #math.equation(block: false, alt: "c squared minus 4 k m less than 0")[$c^(2) − 4 k m < 0$] (system is underdamped).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the Laplace transform solve


      #math.equation(block: true, alt: "m x double prime plus c x prime plus k x equals 0 , x open parenthesis 0 close parenthesis equals a , x prime open parenthesis 0 close parenthesis equals b .")[$m x^(″) + c x^(′) + k x = 0 , #h(1em) x ( 0 ) = a , #h(1em) x^(′) ( 0 ) = b .$]


      where #math.equation(block: false, alt: "m greater than 0 , c greater than 0 , k greater than 0")[$m > 0 , c > 0 , k > 0$], and #math.equation(block: false, alt: "c squared equals 4 k m")[$c^(2) = 4 k m$] (system is critically damped).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x double prime plus x equals u open parenthesis t minus 1 close parenthesis")[$x^(″) + x = u ( t − 1 )$] for initial conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 0")[$x ( 0 ) = 0$] and #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Show the differentiation of the transform property. Suppose #math.equation(block: false, alt: "ℒ { f open parenthesis t close parenthesis } equals F open parenthesis s close parenthesis")[$ℒ \{ f ( t ) \} = F ( s )$], then show


      #math.equation(block: true, alt: "ℒ { minus t f open parenthesis t close parenthesis } equals F prime open parenthesis s close parenthesis .")[$ℒ \{ − t f ( t ) \} = F^(′) ( s ) .$]


      Hint: Differentiate under the integral sign.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x to the power ‴ plus x equals t cubed u open parenthesis t minus 1 close parenthesis")[$x^(‴) + x = t^(3) u ( t − 1 )$] for initial conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1")[$x ( 0 ) = 1$] and #math.equation(block: false, alt: "x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) ( 0 ) = 0$], #math.equation(block: false, alt: "x double prime open parenthesis 0 close parenthesis equals 0")[$x^(″) ( 0 ) = 0$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Show the second shifting property: #math.equation(block: false, alt: "ℒ { f open parenthesis t minus a close parenthesis u open parenthesis t minus a close parenthesis } equals e to the power minus a s ℒ { f open parenthesis t close parenthesis }")[$ℒ \{ f ( t − a ) u ( t − a ) \} = e^(− a s) ℒ \{ f ( t ) \}$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let us think of the mass-spring system with a rocket from Example 6.2.2. We noticed that the solution kept oscillating after the rocket stopped running. The amplitude of the oscillation depends on the time that the rocket was fired (for 4 seconds in the example).


      + Find a formula for the amplitude of the resulting oscillation in terms of the amount of time the rocket is fired.
+ Is there a nonzero time (if so what is it?) for which the rocket fires and the resulting oscillation has amplitude 0 (the mass is not moving)?
]


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


      #math.equation(block: true, alt: "open parenthesis t minus 1 close parenthesis squared i f 1 less than or equal to t less than 2 ,; 3 minus t i f 2 less than or equal to t less than 3 ,; 0 o t h e r w i s e .")[$( t − 1 )^(2) i f " " 1 ≤ t < 2 , \ 3 − t i f " " 2 ≤ t < 3 , \ 0 o t h e r w i s e .$]


      + Sketch the graph of #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$].
+ Write down #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] using the Heaviside function.
+ Solve #math.equation(block: false, alt: "x double prime plus x equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$x^(″) + x = f ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0$] using Laplace transform.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the transfer function for #math.equation(block: false, alt: "m x double prime plus c x prime plus k x equals f open parenthesis t close parenthesis")[$m x^(″) + c x^(′) + k x = f ( t )$] (assuming the initial conditions are zero).
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Using the Heaviside function #math.equation(block: false, alt: "u open parenthesis t close parenthesis")[$u ( t )$], write down the function


      #math.equation(block: true, alt: "0 i f t less than 1 ,; t minus 1 i f 1 less than or equal to t less than 2 ,; i f 2 less than or equal to t .")[$0 i f " " " " " " " " " " t < 1 , \ t − 1 i f " " 1 ≤ t < 2 , \ i f " " " " " " " " " " 2 ≤ t .$]


      Answer


      #math.equation(block: true, alt: "f open parenthesis t close parenthesis equals open parenthesis t minus 1 close parenthesis open parenthesis u open parenthesis t minus 1 close parenthesis minus u open parenthesis t minus 2 close parenthesis close parenthesis plus u open parenthesis t minus 2 close parenthesis")[$f ( t ) = ( t − 1 ) ( u ( t − 1 ) − u ( t − 2 ) ) + u ( t − 2 )$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Solve #math.equation(block: false, alt: "x double prime minus x equals open parenthesis t squared minus 1 close parenthesis u open parenthesis t minus 1 close parenthesis")[$x^(″) − x = ( t^(2) − 1 ) u ( t − 1 )$] for initial conditions #math.equation(block: false, alt: "x open parenthesis 0 close parenthesis equals 1 , x prime open parenthesis 0 close parenthesis equals 2")[$x ( 0 ) = 1 , x^(′) ( 0 ) = 2$] using the Laplace transform.


      Answer


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


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the transfer function for #math.equation(block: false, alt: "x prime plus x equals f open parenthesis t close parenthesis")[$x^(′) + x = f ( t )$] (assuming the initial conditions are zero).


      Answer


      #math.equation(block: true, alt: "H open parenthesis s close parenthesis equals the fraction 1 over s plus 1")[$H ( s ) = frac(1, s + 1)$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/6%3A_The_Laplace_Transform/6.3%3A_Convolution")[6.3: Convolution]
    


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t squared")[$f ( t ) = t^(2)$] for #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$], and #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals u open parenthesis t minus 1 close parenthesis")[$g ( t ) = u ( t − 1 )$]. Compute #math.equation(block: false, alt: "f * g")[$f * g$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals t")[$f ( t ) = t$] for #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$], and #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals sin t")[$g ( t ) = sin t$] for #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$]. Compute #math.equation(block: false, alt: "f * g")[$f * g$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the solution to


      #math.equation(block: true, alt: "m x double prime plus c x prime plus k x equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 ,")[$m x^(″) + c x^(′) + k x = f ( t ) , #h(1em) x ( 0 ) = 0 , #h(1em) x^(′) ( 0 ) = 0 ,$]


      for an arbitrary function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], where #math.equation(block: false, alt: "m greater than 0 , c greater than 0 , k greater than 0")[$m > 0 , c > 0 , k > 0$], and #math.equation(block: false, alt: "c squared minus 4 k m greater than 0")[$c^(2) − 4 k m > 0$] (system is overdamped). Write the solution as a definite integral.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the solution to


      #math.equation(block: true, alt: "m x double prime plus c x prime plus k x equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 ,")[$m x^(″) + c x^(′) + k x = f ( t ) , #h(1em) x ( 0 ) = 0 , #h(1em) x^(′) ( 0 ) = 0 ,$]


      for an arbitrary function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], where #math.equation(block: false, alt: "m greater than 0 , c greater than 0 , k greater than 0")[$m > 0 , c > 0 , k > 0$], and #math.equation(block: false, alt: "c squared minus 4 k m less than 0")[$c^(2) − 4 k m < 0$] (system is underdamped). Write the solution as a definite integral.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Find the solution to


      #math.equation(block: true, alt: "m x double prime plus c x prime plus k x equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 ,")[$m x^(″) + c x^(′) + k x = f ( t ) , #h(1em) x ( 0 ) = 0 , #h(1em) x^(′) ( 0 ) = 0 ,$]


      for an arbitrary function #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$], where #math.equation(block: false, alt: "m greater than 0 , c greater than 0 , k greater than 0")[$m > 0 , c > 0 , k > 0$], and #math.equation(block: false, alt: "c squared equals 4 k m")[$c^(2) = 4 k m$] (system is critically damped). Write the solution as a definite integral.
]


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


      #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals e to the power minus t plus ∫ 0 t cos open parenthesis t minus τ close parenthesis x open parenthesis τ close parenthesis d τ .")[$x ( t ) = e^(− t) + ∫_(0)^(t) cos ( t − τ ) x ( τ ) " " d τ .$]
]


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


      #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals cos t plus ∫ 0 t cos open parenthesis t minus τ close parenthesis x open parenthesis τ close parenthesis d τ .")[$x ( t ) = cos t + ∫_(0)^(t) cos ( t − τ ) x ( τ ) " " d τ .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Compute #math.equation(block: false, alt: "ℒ to the power minus 1 { the fraction s over open parenthesis s squared plus 4 close parenthesis squared }")[$ℒ^(− 1) \{ frac(s, ( s^(2) + 4 )^(2)) \}$] using convolution.
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Write down the solution to #math.equation(block: false, alt: "x double prime minus 2 x equals e to the power minus t squared , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$x^(″) − 2 x = e^(− t^(2)) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0$] as a definite integral. Hint: Do not try to compute the Laplace transform of #math.equation(block: false, alt: "e to the power minus t squared")[$e^(− t^(2))$].
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Let #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals cos t")[$f ( t ) = cos t$] for #math.equation(block: false, alt: "t greater than or equal to 0")[$t ≥ 0$], and #math.equation(block: false, alt: "g open parenthesis t close parenthesis equals e to the power minus t")[$g ( t ) = e^(− t)$]. Compute #math.equation(block: false, alt: "f * g")[$f * g$].


      Answer


      #math.equation(block: true, alt: "the fraction 1 over 2 open parenthesis cos t plus sin t minus e to the power minus t close parenthesis")[$frac(1, 2) ( cos t + sin t − e^(− t) )$]
]


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Compute #math.equation(block: false, alt: "ℒ to the power minus 1 { the fraction 5 over s to the power 4 plus s squared }")[$ℒ^(− 1) \{ frac(5, s^(4) + s^(2)) \}$] using convolution.


      Answer


      #math.equation(block: true, alt: "5 t minus 5 sin t")[$5 t − 5 sin t$]
]


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Solve #math.equation(block: false, alt: "x double prime plus x equals sin t , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$x^(″) + x = sin t , x ( 0 ) = 0 , x^(′) ( 0 ) = 0$] using convolution.


      Answer


      #math.equation(block: true, alt: "the fraction 1 over 2 open parenthesis sin t minus t cos t close parenthesis")[$frac(1, 2) ( sin t − t cos t )$]
]


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Solve #math.equation(block: false, alt: "x to the power ‴ plus x prime equals f open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 , x double prime open parenthesis 0 close parenthesis equals 0")[$x^(‴) + x^(′) = f ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0 , x^(″) ( 0 ) = 0$] using convolution. Write the result as a definite integral.


      Answer


      #math.equation(block: true, alt: "∫ 0 t f open parenthesis τ close parenthesis open parenthesis 1 minus cos open parenthesis t minus τ close parenthesis close parenthesis d τ")[$∫_(0)^(t) f ( τ ) ( 1 − cos ( t − τ ) ) d τ$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/6%3A_The_Laplace_Transform/6.4%3A_Dirac_Delta_and_Impulse_Response")[6.4: Dirac delta and impulse response]
    


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Solve (find the impulse response) #math.equation(block: false, alt: "x double prime plus x prime plus x equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 .")[$x^(″) + x^(′) + x = δ ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0 .$]
]


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Solve (find the impulse response) #math.equation(block: false, alt: "x double prime plus 2 x prime plus x equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 .")[$x^(″) + 2 x^(′) + x = δ ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0 .$]
]


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A pulse can come later and can be bigger. Solve #math.equation(block: false, alt: "x double prime plus 4 x equals 4 δ open parenthesis t minus 1 close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 .")[$x^(″) + 4 x = 4 δ ( t − 1 ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0 .$]
]


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Suppose that #math.equation(block: false, alt: "f open parenthesis t close parenthesis")[$f ( t )$] and #math.equation(block: false, alt: "g open parenthesis t close parenthesis")[$g ( t )$] are differentiable functions and suppose that #math.equation(block: false, alt: "f open parenthesis t close parenthesis equals g open parenthesis t close parenthesis equals 0")[$f ( t ) = g ( t ) = 0$] for all #math.equation(block: false, alt: "t less than or equal to 0")[$t ≤ 0$]. Show that #math.equation(block: true, alt: "open parenthesis f * g close parenthesis prime open parenthesis t close parenthesis equals open parenthesis f prime * g close parenthesis open parenthesis t close parenthesis equals open parenthesis f * g prime close parenthesis open parenthesis t close parenthesis .")[$( f * g )^(′) ( t ) = ( f^(′) * g ) ( t ) = ( f * g^(′) ) ( t ) .$]
]


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Suppose that #math.equation(block: false, alt: "L x equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$L x = δ ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0$], has the solution #math.equation(block: false, alt: "x equals e to the power minus t")[$x = e^(− t)$] for #math.equation(block: false, alt: "t greater than 0")[$t > 0$]. Find the solution to #math.equation(block: false, alt: "L x equals t squared , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$L x = t^(2) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0$] for #math.equation(block: false, alt: "t greater than 0")[$t > 0$].
]


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Compute #math.equation(block: false, alt: "ℒ to the power minus 1 { the fraction s squared plus s plus 1 over s squared }")[$ℒ^(− 1) \{ frac(s^(2) + s + 1, s^(2)) \}$].
]


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#emph[(challenging)]


      Solve Example 6.4.3 via integrating 4 times in the #math.equation(block: false, alt: "x")[$x$] variable.
]


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Suppose we have a beam of length #math.equation(block: false, alt: "1")[$1$] simply supported at the ends and suppose that force #math.equation(block: false, alt: "F equals 1")[$F = 1$] is applied at #math.equation(block: false, alt: "x equals the fraction 3 over 4")[$x = frac(3, 4)$] in the downward direction. Suppose that #math.equation(block: false, alt: "E I equals 1")[$E I = 1$] for simplicity. Find the beam deflection #math.equation(block: false, alt: "y open parenthesis x close parenthesis")[$y ( x )$].
]


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Solve (find the impulse response) #math.equation(block: false, alt: "x double prime equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$x^(″) = δ ( t ) , #h(0.222em) x ( 0 ) = 0 , #h(0.222em) x^(′) ( 0 ) = 0$].


      Answer


      #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals t")[$x ( t ) = t$]
]


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Solve (find the impulse response) #math.equation(block: false, alt: "x prime plus a x equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$x^(′) + a x = δ ( t ) , #h(0.222em) x ( 0 ) = 0 , #h(0.222em) x^(′) ( 0 ) = 0$].


      Answer


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


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Suppose that #math.equation(block: false, alt: "L x equals δ open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0")[$L x = δ ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0$], has the solution #math.equation(block: false, alt: "x open parenthesis t close parenthesis equals cos open parenthesis t close parenthesis")[$x ( t ) = cos ( t )$] for #math.equation(block: false, alt: "t greater than 0")[$t > 0$]. Find (in closed form) the solution to #math.equation(block: false, alt: "L x equals sin open parenthesis t close parenthesis , x open parenthesis 0 close parenthesis equals 0 , x prime open parenthesis 0 close parenthesis equals 0 f o r t greater than 0")[$L x = sin ( t ) , x ( 0 ) = 0 , x^(′) ( 0 ) = 0 f o r t > 0$].


      Answer


      #math.equation(block: true, alt: "x open parenthesis t close parenthesis equals open parenthesis cos * sin close parenthesis open parenthesis t close parenthesis equals the fraction 1 over 2 t sin open parenthesis t close parenthesis")[$x ( t ) = ( cos * sin ) ( t ) = frac(1, 2) t sin ( t )$]
]


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Compute #math.equation(block: false, alt: "ℒ to the power minus 1 { the fraction s squared over s squared plus 1 }")[$ℒ^(− 1) \{ frac(s^(2), s^(2) + 1) \}$].


      Answer


      #math.equation(block: true, alt: "δ open parenthesis t close parenthesis minus sin open parenthesis t close parenthesis")[$δ ( t ) − sin ( t )$]
]


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Compute #math.equation(block: false, alt: "ℒ to the power minus 1 { the fraction 3 s squared e to the power minus s plus 2 over s squared }")[$ℒ^(− 1) \{ frac(3 s^(2) e^(− s) + 2, s^(2)) \}$].


      Answer


      #math.equation(block: true, alt: "3 δ open parenthesis t minus 1 close parenthesis plus 2 t")[$3 δ ( t − 1 ) + 2 t$]
]


    
=== 
      #link("https://math.libretexts.org/Bookshelves/Differential_Equations/Differential_Equations_for_Engineers_(Lebl)/6%3A_The_Laplace_Transform/6.5%3A_Solving_PDEs_with_the_Laplace_Transform")[6.5: Solving PDEs with the Laplace Transform]
    


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Solve #math.equation(block: true, alt: "y sub t plus y sub x equals 1 , 0 less than x less than infinity , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals 1 , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t) + y_(x) = 1 , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y ( 0 , t ) = 1 , #h(1em) y ( x , 0 ) = 0 .$]
]


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Solve #math.equation(block: true, alt: "y sub t plus α y sub x equals 0 , 0 less than x less than infinity , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals t , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t) + α y_(x) = 0 , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y ( 0 , t ) = t , #h(1em) y ( x , 0 ) = 0 .$]
]


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Solve #math.equation(block: true, alt: "y sub t plus 2 y sub x equals x plus t , 0 less than x less than infinity , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t) + 2 y_(x) = x + t , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y ( 0 , t ) = 0 , #h(1em) y ( x , 0 ) = 0 .$]
]


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For an #math.equation(block: false, alt: "α greater than 0")[$α > 0$], solve #math.equation(block: true, alt: "y sub t plus α y sub x plus y equals 0 , 0 less than x less than infinity , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals sin open parenthesis t close parenthesis , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t) + α y_(x) + y = 0 , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y ( 0 , t ) = sin ( t ) , #h(1em) y ( x , 0 ) = 0 .$]
]


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Find the corresponding ODE problem for #math.equation(block: false, alt: "Y open parenthesis x close parenthesis")[$Y ( x )$], after transforming the #math.equation(block: false, alt: "t")[$t$] variable #math.equation(block: true, alt: "y sub t t plus 3 y sub x x plus y sub x t plus 3 y sub x plus y equals sin open parenthesis x close parenthesis plus t , 0 less than x less than 1 , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals 1 , y open parenthesis 1 , t close parenthesis equals t , y open parenthesis x , 0 close parenthesis equals 1 minus x , y sub t open parenthesis x , 0 close parenthesis equals 1 .")[$y_(t t) + 3 y_(x x) + y_(x t) + 3 y_(x) + y = sin ( x ) + t , #h(2em) 0 < x < 1 , #h(0.5em) t > 0 , \ y ( 0 , t ) = 1 , #h(1em) y ( 1 , t ) = t , #h(1em) y ( x , 0 ) = 1 − x , #h(1em) y_(t) ( x , 0 ) = 1 .$] Do not solve the problem.
]


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Write down a solution to #math.equation(block: true, alt: "y sub t equals y sub x x , 0 less than x less than infinity , t greater than 0 ,; y sub x open parenthesis 0 , t close parenthesis equals e to the power minus t , y open parenthesis x , 0 close parenthesis equals 0 ,")[$y_(t) = y_(x x) , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y_(x) ( 0 , t ) = e^(− t) , #h(1em) y ( x , 0 ) = 0 ,$] as an definite integral (convolution).
]


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Use the Laplace transform in #math.equation(block: false, alt: "t")[$t$] to solve #math.equation(block: true, alt: "y sub t t equals y sub x x , minus infinity less than x less than infinity , t greater than 0 ,; y sub t open parenthesis x , 0 close parenthesis equals sin open parenthesis x close parenthesis , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t t) = y_(x x) , #h(2em) − ∞ < x < ∞ , #h(0.5em) t > 0 , \ y_(t) ( x , 0 ) = sin ( x ) , #h(1em) y ( x , 0 ) = 0 .$] Hint: Note that #math.equation(block: false, alt: "e to the power s x")[$e^(s x)$] does not go to zero as #math.equation(block: false, alt: "s → infinity")[$s → ∞$] for positive #math.equation(block: false, alt: "x")[$x$], and #math.equation(block: false, alt: "e to the power minus s x")[$e^(− s x)$] does not go to zero as #math.equation(block: false, alt: "s → infinity")[$s → ∞$] for negative #math.equation(block: false, alt: "x")[$x$].
]


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Solve #math.equation(block: true, alt: "y sub t plus y sub x equals 1 , 0 less than x less than infinity , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t) + y_(x) = 1 , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y ( 0 , t ) = 0 , #h(1em) y ( x , 0 ) = 0 .$]


      Answer


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


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For a #math.equation(block: false, alt: "c greater than 0")[$c > 0$], solve #math.equation(block: true, alt: "y sub t plus y sub x plus c y equals 0 , 0 less than x less than infinity , t greater than 0 ,; y open parenthesis 0 , t close parenthesis equals sin open parenthesis t close parenthesis , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t) + y_(x) + c y = 0 , #h(2em) 0 < x < ∞ , #h(0.5em) t > 0 , \ y ( 0 , t ) = sin ( t ) , #h(1em) y ( x , 0 ) = 0 .$]


      Answer


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


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Find the corresponding ODE problem for #math.equation(block: false, alt: "Y open parenthesis x close parenthesis")[$Y ( x )$], after transforming the #math.equation(block: false, alt: "t")[$t$] variable #math.equation(block: true, alt: "y sub t t plus 3 y sub x x plus y equals x plus t , minus 1 less than x less than 1 , t greater than 0 ,; y open parenthesis minus 1 , t close parenthesis equals 0 , y open parenthesis 1 , t close parenthesis equals 0 , y open parenthesis x , 0 close parenthesis equals open parenthesis 1 minus x squared close parenthesis , y sub t open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t t) + 3 y_(x x) + y = x + t , #h(2em) − 1 < x < 1 , #h(0.5em) t > 0 , \ y ( − 1 , t ) = 0 , #h(1em) y ( 1 , t ) = 0 , #h(1em) y ( x , 0 ) = ( 1 − x^(2) ) , #h(1em) y_(t) ( x , 0 ) = 0 .$] Do not solve the problem.


      Answer


      #math.equation(block: true, alt: "s squared Y open parenthesis x close parenthesis minus s open parenthesis 1 plus x squared close parenthesis plus 3 Y double prime open parenthesis x close parenthesis plus Y open parenthesis x close parenthesis equals the fraction x over s plus the fraction 1 over s squared , Y open parenthesis minus 1 close parenthesis equals 0 , Y open parenthesis 1 close parenthesis equals 0 .")[$s^(2) Y ( x ) − s ( 1 + x^(2) ) + 3 Y^(″) ( x ) + Y ( x ) = frac(x, s) + frac(1, s^(2)) , #h(1em) Y ( − 1 ) = 0 , #h(1em) Y ( 1 ) = 0 .$]
]


    #notebox("Your Turn", rgb("#8a94a6"), rgb("#556666"), rgb("#f7f8fa"))[
Use the Laplace transform in #math.equation(block: false, alt: "t")[$t$] to solve #math.equation(block: true, alt: "y sub t t equals y sub x x , minus infinity less than x less than infinity , t greater than 0 ,; y sub t open parenthesis x , 0 close parenthesis equals x squared , y open parenthesis x , 0 close parenthesis equals 0 .")[$y_(t t) = y_(x x) , #h(2em) − ∞ < x < ∞ , #h(0.5em) t > 0 , \ y_(t) ( x , 0 ) = x^(2) , #h(1em) y ( x , 0 ) = 0 .$] Hint: Note that #math.equation(block: false, alt: "e to the power s x")[$e^(s x)$] does not go to zero as #math.equation(block: false, alt: "s → infinity")[$s → ∞$] for positive #math.equation(block: false, alt: "x")[$x$], and #math.equation(block: false, alt: "e to the power minus s x")[$e^(− s x)$] does not go to zero as #math.equation(block: false, alt: "s → infinity")[$s → ∞$] for negative #math.equation(block: false, alt: "x")[$x$].


      Answer


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