Jun's question via email about Laplace Transform

Prove It

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$\displaystyle y\left( t \right)$ satisfies the initial value problem

$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 4\,y= -8\,H\left( t - 6 \right) , \quad y\left( 0 \right) = 2 , \,\, y'\left( 0 \right) = 0$

Find the solution to the initial value problem using Laplace Transforms.
Upon taking the Laplace Transform of the equation we have

\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &= -\frac{8\,\mathrm{e}^{-6\,s}}{s} \\ s^2 \,Y\left( s \right) - 2\,s - 0 + 4\,Y\left( s \right) &= -\frac{8\,\mathrm{e}^{-6\,s}}{s}\\ \left( s^2 + 4 \right) Y\left( s \right) - 2\,s &= -\frac{8\,\mathrm{e}^{-6\,s}}{s} \\ \left( s^2 + 4\right) Y\left( s \right) &= 2\,s - \frac{8\,\mathrm{e}^{-6\,s}}{s} \\ Y\left( s \right) &= \frac{2\,s}{s^2 + 4} - \frac{8\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } \\ Y\left( s \right) &= 2 \left[ \frac{s}{s^2 + 4} - \frac{4\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } \right] \end{align*}

The first term's Inverse Transform can be read off the tables. The second requires the second shift theorem: $\displaystyle \mathcal{L}\,\left\{ f\left( t - a \right) \, H\left( t - a \right) \right\} = \mathrm{e}^{-a\,s}\,F\left( s \right)$.

$\displaystyle F\left( s \right) = \frac{4}{s\left( s^2 + 4 \right) }$

Applying Partial Fractions:

\displaystyle \begin{align*} \frac{A}{s} + \frac{B\,s + C}{s^2 + 4} &\equiv \frac{4}{s\left( s^2 + 4 \right) } \\ A\left( s^2 + 4 \right) + \left( B\,s + C \right) s &\equiv 4 \end{align*}

Let $\displaystyle s = 0 \implies 4\,A = 4 \implies A = 1$, then

\displaystyle \begin{align*} 1\left( s^2 + 4 \right) + \left( B\,s + C \right) s &\equiv 4 \\ s^2 + 4 + B\,s^2 + C\,s &\equiv 4 \\ \left( B + 1 \right) s^2 + C\,s + 4 &\equiv 0\,s^2 + 0\,s + 4 \end{align*}

It's clear that $\displaystyle B + 1 = 0 \implies B = -1$ and $\displaystyle C = 0$. Thus

\displaystyle \begin{align*} F\left( s \right) &= \frac{1}{s} - \frac{s}{s^2 + 4} \\ f\left( t \right) &= 1 - \cos{ \left( 2\,t \right) } \\ f\left( t - 6 \right) \, H\left( t - 6 \right) &= \left\{ 1 - \cos{ \left[ 2 \left( t - 6 \right) \right] } \right\} \, H\left( t - 6 \right) \end{align*}

So from our original DE

\displaystyle \begin{align*} Y\left( s \right) &= 2 \left[ \frac{s}{s^2 + 4} - \frac{4\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } \right] \\ \\ y \left( t \right) &= 2\left[ \cos{ \left( 2\,t \right) } - \left\{ 1 - \cos{ \left[ 2\left( t - 6 \right) \right] } \right\} \, H\left( t - 6 \right) \right] \\ &= 2 \left[ \cos{ \left( 2\,t \right) } + \left\{ \cos{ \left[ 2\left( t - 6 \right) \right] } - 1 \right\} \, H\left( t - 6 \right) \right] \end{align*}