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Massaad's question via email about Laplace Transforms

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
$\displaystyle y\left( t \right)$ satisfies the initial value problem:

$\displaystyle \frac{\mathrm{d}^2y}{\mathrm{d}x^2} - 5\,\frac{\mathrm{d}y}{\mathrm{d}t} - 6\,y = -126\,H\left( t - 6 \right) , \quad y\left( 0 \right) = -5, \,\, y'\left( 0 \right) = 5$

Find the solution to the initial value problem using Laplace Transforms.
Taking the Laplace Transform of the equation gives

$\displaystyle \begin{align*} s^2\,Y\left( s \right) - s\,y\left( 0 \right) - y'\left( 0 \right) - 5\left[ s\,Y\left( s \right) - y\left( 0 \right) \right] - 6\,Y\left( s \right) &= -\frac{126\,\mathrm{e}^{-6\,s}}{s} \\
s^2\,Y\left( s \right) - s\left( -5 \right) - 5 - 5 \left[ s\,Y\left( s \right) - \left( -5 \right) \right] - 6\,Y\left( s \right) &= -\frac{126\,\mathrm{e}^{-6\,s}}{s} \\
s^2\,Y\left( s \right) + 5\,s - 5 - 5\,s\,Y\left( s \right) - 25 - 6\,Y\left( s \right) &= -\frac{126\,\mathrm{e}^{-6\,s}}{s} \\
\left( s^2 - 5\,s - 6 \right) Y\left( s \right) + 5\,s - 30 &= -\frac{126\,\mathrm{e}^{-6\,s}}{s} \\
\left( s - 6 \right) \left( s + 1 \right) Y\left( s \right) + 5 \left( s - 6 \right) &= -\frac{126\,\mathrm{e}^{-6\,s}}{s} \\
\left( s + 1 \right) Y\left( s \right) + 5 &= -\frac{126\,\mathrm{e}^{-6\,s}}{s\left( s - 6 \right) } \\
\left( s + 1 \right) Y\left( s \right) &= -\frac{126\,\mathrm{e}^{-6\,s}}{s\left( s - 6 \right) } - 5 \\
Y\left( s \right) &= -\frac{126\,\mathrm{e}^{-6\,s}}{s\left( s - 6 \right) \left( s + 1 \right) } - \frac{5}{s + 1}\end{align*}$

The second term is easy to find the inverse transform of: $\displaystyle \mathcal{L}^{-1}\,\left\{ \frac{5}{s + 1} \right\} = 5\,\mathrm{e}^{-t} $.

For the first term, due to the exponential function, it suggests a second shift: $\displaystyle \mathcal{L}\,\left\{ f\left( t - a \right) \,H\left( t - a \right) \right\} = \mathrm{e}^{-a\,s}\,F\left( s \right) $.

So in this case, we have $\displaystyle F\left( s \right) = -\frac{126}{s\left( s - 6 \right) \left( s + 1 \right) } $.

To find $\displaystyle f\left( t \right) $ we will need partial fractions:

$\displaystyle \begin{align*} \frac{A}{s} + \frac{B}{s - 6} + \frac{C}{s + 1} &\equiv \frac{-126}{s\left( s - 6 \right) \left( s + 1 \right) } \\
A \left( s - 6 \right) \left( s + 1 \right) + B\,s\left( s + 1 \right) + C\,s \left( s - 6 \right) &\equiv -126 \end{align*} $

Let $\displaystyle s = 0 \implies -6\,A = -126 \implies A = 21$

Let $\displaystyle s = 6 \implies 42\,B = -126 \implies B = -3$

Let $\displaystyle s = -1 \implies 7\,C = -126 \implies C = -18 $

So

$\displaystyle \begin{align*} F\left( s \right) &= 21 \left( \frac{1}{s} \right) - 3 \left( \frac{1}{s - 6} \right) - 18 \left( \frac{1}{s + 1} \right) \\
f\left( t \right) &= 21 - 3\,\mathrm{e}^{6\,t} - 18\,\mathrm{e}^{-t} \end{align*}$

Thus $\displaystyle \mathcal{L}^{-1}\,\left\{ -\frac{126\,\mathrm{e}^{-6\,s}}{s\left( s - 6 \right) \left( s + 1 \right) } \right\} = \left[ 21 - 3\,\mathrm{e}^{6\,\left( t - 6 \right) } - 18\,\mathrm{e}^{-\left( t - 6 \right) } \right]\,H\left( t - 6 \right) $ by the second shift theorem.

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

$\displaystyle y\left( t \right) = \left[ 21 - 3\,\mathrm{e}^{6\,\left( t - 6 \right) } - 18\,\mathrm{e}^{-\left( t - 6 \right) } \right] \, H\left( t - 6 \right) - 5\,\mathrm{e}^{-t}$