Collin's question via email about solving a DE using Laplace Transforms

[Prove It]

Solve the following Differential Equation:

$\displaystyle \begin{align*} y'' + 4\,y = \mathrm{H}\,\left( t - 7 \right) \textrm{ with } y\left( 0 \right) = 0 \textrm{ and } y'\left( 0 \right) = -10 \end{align*}$

Taking the Laplace Transform of both sides we have

$\displaystyle \begin{align*} \mathcal{L}\,\left\{ y'' + 4\,y \right\} &= \mathcal{L}\,\left\{ \mathrm{H}\,\left( t - 7 \right) \right\} \\ s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 4\,Y\left( s \right) &= \frac{\mathrm{e}^{-7\,s}}{s} \\ s^2\,Y\left( s \right) + 10 + 4\,Y\left( s \right) &= \frac{ \mathrm{e}^{-7\,s}}{s} \\ \left( s^2 + 4 \right) \, Y\left( s \right) &= \frac{\mathrm{e}^{-7\,s}}{s} - 10 \\ Y\left( s \right) &= \frac{\mathrm{e}^{-7\,s}}{s\,\left( s^2 + 4 \right) } - \frac{10}{s^2 + 4} \end{align*}$

So to solve the DE, all that we require now is to take the Inverse Laplace Transform of this result. To do this with the first term, we will need to use Partial Fractions:

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

Let $\displaystyle \begin{align*} s = 0 \end{align*}$ to find $\displaystyle \begin{align*} 4\,A = 1 \implies A = \frac{1}{4} \end{align*}$.

The coefficient of $\displaystyle \begin{align*} s^2 \end{align*}$ is $\displaystyle \begin{align*} A + B \end{align*}$ on the left and $\displaystyle \begin{align*} 0 \end{align*}$ on the right, and as $\displaystyle \begin{align*} A = \frac{1}{4} \end{align*}$ we have $\displaystyle \begin{align*} \frac{1}{4} + B = 0 \implies B = -\frac{1}{4} \end{align*}$.

The coefficient of $\displaystyle \begin{align*} s \end{align*}$ is $\displaystyle \begin{align*} C \end{align*}$ on the left and $\displaystyle \begin{align*} 0 \end{align*}$ on the right, thus $\displaystyle \begin{align*} C = 0 \end{align*}$.

Therefore $\displaystyle \begin{align*} \frac{1}{s\,\left( s^2 + 4 \right) } = \frac{1}{4\,s} - \frac{s}{4\,\left( s^2 + 4 \right) } \end{align*}$. So that means

$\displaystyle \begin{align*} y &= \mathcal{L}^{-1}\,\left\{ \frac{\mathrm{e}^{-7\,s}}{s\,\left( s^2 + 4 \right) } - \frac{10}{s^2 + 4} \right\} \\ &= \frac{1}{4} \mathcal{L}^{-1}\,\left\{ \frac{\mathrm{e}^{-7\,s}}{s} \right\} - \frac{1}{8} \,\mathcal{L}^{-1}\,\left\{ \mathrm{e}^{-7\,s} \,\left( \frac{2}{ s^2 + 2^2 } \right) \right\} - 5\,\mathcal{L}^{-1}\,\left\{ \frac{2}{s^2 + 2^2} \right\} \end{align*}$

The first and third terms are very straightforward. For the second term, we need to make use of the rule $\displaystyle \begin{align*} \mathcal{L}\,\left\{ f\left( t - c \right) \, \mathrm{H}\,\left( t - c \right) \right\} = \mathrm{e}^{-c\,s}\,F\left( s \right) \end{align*}$. We can read off that $\displaystyle \begin{align*} F\left( s \right) = \frac{2}{ s^2 + 2^2} \end{align*}$ and thus $\displaystyle \begin{align*} f\left( t \right) = \sin{ \left( 2\,t \right) } \end{align*}$. From here we can get that $\displaystyle \begin{align*} f\left( t - 7 \right) = \sin{ \left[ 2\,\left( t - 7 \right) \right] } \end{align*}$, and finally the solution to the DE is

$\displaystyle \begin{align*} y = \mathrm{H}\,\left( t - 7 \right) - \frac{1}{8}\sin{ \left[ 2\,\left( t - 7 \right) \right] }\,\mathrm{H}\,\left( t - 7 \right) - 5 \sin{ \left( 2\,t \right) } \end{align*}$.

 

[Mark44]

Careful students will check the solution to ensure that it actually is a solution to the DE.

 

[pasmith]

By using the inversion formula and residue calculus I get $$\begin{split} y(t) &= 10 \left( \frac{e^{2it}}{4i} + \frac{e^{-2it}}{-4i}\right) + \left( \frac14 + \frac{e^{2i(t-7)}}{(2i)(4i)} + \frac{e^{-2i(t-7)}}{(-2i)(-4i)} \right)H(t-7) \\ &= 5\sin(2t) + \left(\tfrac14 - \tfrac14\cos(2(t-7))\right)H(t-7).\end{split}$$ I feel this involved less work, with less scope for error, than manipulating $\dfrac{10}{s^2 + 4} + \dfrac{e^{-7s}}{s(s^2+4)}$ into a form where its inverse transform can be read from tables.