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

#### Prove It

##### Well-known member
MHB Math Helper
Solve $\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 7\,\frac{\mathrm{d}y}{\mathrm{d}t} + 6\,y = 60\,H\left( t - 6 \right) , \quad y\left( 0 \right) = -4, \quad y'\left( 0 \right) = 4$ using Laplace Transforms.
Taking the Laplace Transform of the equation gives

\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) + 7 \left[ s\,Y\left( s \right) - y\left( 0\right) \right] + 6\,Y\left( s \right) &= \frac{60\,\mathrm{e}^{-6\,s}}{s} \\ s^2\,Y\left( s \right) - s\left( - 4\right) - 4 + 7\left[ s \,Y\left( s \right) - \left( -4 \right) \right] + 6\,Y\left( s \right) &= \frac{60\,\mathrm{e}^{-6\,s}}{s} \\ s^2\,Y\left( s \right) + 4\,s - 4 + 7\,s\,Y\left( s \right) + 28 + 6\,Y\left( s \right) &= \frac{60\,\mathrm{e}^{-s}}{s} \\ \left( s^2 + 7\,s + 6 \right) Y\left( s \right) + 4\,s + 24 &= \frac{60\,\mathrm{e}^{-6\,s}}{s} \\ \left( s + 1 \right) \left( s + 6 \right) Y\left( s \right) + 4 \left( s + 6 \right) &= \frac{60\,\mathrm{e}^{-6\,s}}{s} \ \left( s + 1 \right) Y\left( s \right) + 4 &= \frac{60\,\mathrm{e}^{-6\,s}}{s\left( s + 6 \right) } \\ \left( s + 1 \right) Y\left( s \right) &= \frac{60\,\mathrm{e}^{-6\,s}}{s \left( s + 6 \right) } - 4 \\ Y\left( s \right) &= \frac{60\,\mathrm{e}^{-6\,s}}{s \left( s + 1 \right) \left( s + 6 \right) } - \frac{4}{s + 1} \end{align*}

Evaluating $\displaystyle y\left( t \right)$ will require taking the Inverse Laplace Transform of this function.

The second term is easy as we can read the inverse transform from the table: $\displaystyle \mathcal{L}^{-1}\,\left\{ \frac{4}{s + 1} \right\} = 4\,\mathrm{e}^{-t}$.

For the first term, the exponential function factor suggests that we need to use the second shift theorem: $\displaystyle \mathcal{L}^{-1}\,\left\{ \mathrm{e}^{-a\,s}\,F\left( s \right) \right\} = f\left( t - a \right) H\left( t - a \right)$.

Here $\displaystyle F\left( s \right) = \frac{60}{s\left( s + 1 \right) \left( s + 6 \right) }$. To find $\displaystyle f\left( t \right)$ we will first need Partial Fractions.

\displaystyle \begin{align*} \frac{A}{s} + \frac{B}{s + 1} + \frac{C}{s + 6} &\equiv \frac{60}{s\left( s + 1 \right) \left( s + 6 \right) } \\ A\left( s + 1 \right) \left( s + 6 \right) + B\,s\left( s + 6 \right) + C\,s\left( s + 1 \right) &\equiv 60 \end{align*}

Let $\displaystyle s = 0 \implies 6\,A = 60 \implies A = 10$.

Let $\displaystyle s = -1 \implies -5\,B = 60 \implies B = -12$.

Let $\displaystyle s = -6 \implies 30\,C = 60 \implies C = 2$. Thus

\displaystyle \begin{align*} F\left( s \right) &= 10\left( \frac{1}{s} \right) - 12\left( \frac{1}{s + 1} \right) + 2\left( \frac{1}{s + 6} \right) \\ \\ f\left( t \right) &= 10 - 12\,\mathrm{e}^{-t} + 2\,\mathrm{e}^{-6} \\ \\ f\left( t - 6 \right) &= 10 - 12\,\mathrm{e}^{-\left( t - 6 \right) } + 2\,\mathrm{e}^{-6\,\left( t - 6 \right) } \\ \\ f\left( t - 6 \right) H\left( t - 6 \right) &= \left[ 10 - 12\,\mathrm{e}^{-\left( t - 6 \right) } + 2\,\mathrm{e}^{-6\,\left( t - 6 \right) } \right] H\left( t - 6 \right) \end{align*}

So we can finally write down the solution to the DE:

$\displaystyle y\left( t \right) = \left[ 10 - 12\,\mathrm{e}^{-\left( t - 6 \right) } + 2\,\mathrm{e}^{-6\,\left( t - 6 \right) } \right] H\left( t - 6 \right) - 4\,\mathrm{e}^{-t}$