Jun's question via email about Laplace Transform

[Prove It]

$\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*} $

 

[benorin]

Was it given that $y’(0)=0$? The second line after you apply the transform (the 0), I think this should be replaced by an arbitrary constant, say c? Try that. Since it’s a second order IVP with only one initial value given the solution should contain an unknown constant.

Here’s you work accounting for this constant up to a certain point with differences $\boxed{\text{new term(s)}}$.

Your work:
Upon taking the Laplace Transform of the equation we have

$ \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 - \boxed{c} + 4\,Y\left( s \right) &=& -\frac{8\,\mathrm{e}^{-6\,s}}{s}+\boxed{c} \\ \left( s^2 + 4 \right) Y\left( s \right) - 2\,s &=& -\frac{8\,\mathrm{e}^{-6\,s}}{s}+\boxed{c} \\ \left( s^2 + 4\right) Y\left( s \right) &=& 2\,s - \frac{8\,\mathrm{e}^{-6\,s}}{s}+\boxed{c} \\ Y\left( s \right) &=& \frac{2\,s}{s^2 + 4} - \frac{8\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } +\boxed{\frac{c}{s^2+4}} \\ Y\left( s \right) &=& 2 \left[ \frac{s}{s^2 + 4} - \frac{4\,\mathrm{e}^{-6\,s}}{s\left( s^2 + 4 \right) } + \boxed{\frac{c}{2(s^2+4)}} \right] \\ \end{align*}$

 

[benorin]

Sorry had latex problems, refresh the page. Simple from there, just take the inverse transform of the last boxed term. Should be $c_1 \sin (2t)$ where $c_1 = \tfrac{c}{4}$. Got it from there?