View source for Exercise: Solving an IVP Problem with Laplace Transforms

==Author==

John Hawkins

==Problem Statement==

Solve the following initial value problem using Laplace Transforms:
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<center>

<math>y^{\prime\prime}-4y=12x,\ \ y(0)=4,\ \ y^\prime(0)=1.</math>

</center>
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Note: This problem was solved by Zill without the use of Laplace Transforms.<ref>Dennis G. Zill, ''A first course in Differential Equations'', 8th ed., Int. ed (Belmont, CA: Thomson Learning, 2005), 128.</ref>
==Solution==
Given the initial ODE
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<center><math>y^{\prime\prime}-4y=12x</math></center>

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we take the Laplace transform of both sides
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<center><math>\mathcal{L}\left\{y^{\prime\prime}-4y\right\}=\mathcal{L}\left\{12x\right\}</math></center>

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Using the transforms displayed in [[Laplace Transform]], we find this to be
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<center><math>[s^2Y(s)-sy(0)-y^\prime(0)]-4Y(s)=\frac{12}{s^2}</math></center>

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which, with initial values substituted, gives
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<center><math>\Rightarrow (s^2-4)Y(s)-4s-1=\frac{12}{s^2}</math></center>

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Hence,
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<center>

<math>Y(s)=\frac{\frac{12}{s^2}+4s+1}{s^2-4}</math>

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::<math>=\frac{12+4s^3+s^2}{s^4-4s^2}</math>

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:::<math>=\frac{12+s^2+4s^3}{s^2(s-2)(s+2)}</math>

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Using a calculator to expand this, we have

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<center><math>Y(s)=\frac{1}{s+2}+\frac{3}{s-2}-\frac{3}{s^2}</math></center>

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And therefore, using the equations on [[Laplace Transform]] to perform an inverse Laplace transform, we have our solution:
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<center><math>\ y(x)=e^{-2x}+3e^{2x}-3x</math></center>

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This equation matches that found by Zill, providing confirmation of a correct solution.

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