Solving linear ODEs using the Laplace transform

Problem: Solve the ODE $\;y''+4y=0\;$ with initial conditions $\;y(0)=0,\;y'(0)=1$

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Solution

Transform the ODE

Using Laplace transform properties: \[ \mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0),\quad \mathcal{L}\{y\}=Y(s). \] So \[ s^2Y(s)-sy(0)-y'(0)+4Y(s)=0. \]

Apply initial conditions

With $y(0)=0$ and $y'(0)=1$: \[ s^2Y(s)-0-1+4Y(s)=0 \;\Rightarrow\; (s^2+4)Y(s)=1. \]

Solve for $Y(s)$

\[ Y(s)=\frac{1}{s^2+4}. \]

Rewrite for inverse transform

\[ Y(s)=\frac{1}{2}\cdot\frac{2}{s^2+2^2}. \]

Inverse transform

Using $\mathcal{L}^{-1}\{\tfrac{a}{s^2+a^2}\}=\sin(at)$: \[ y(t)=\frac{1}{2}\sin(2t). \]



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