MHB Mahesh's question via email about Laplace Transforms (2)

[Prove It]

$\displaystyle f\left( t \right)$ satisfies the integral equation

$\displaystyle f\left( t \right) = 7\,t - 3\int_0^t{ f\left( u \right) \,\mathrm{e}^{-3\,\left( t - u \right) } \,\mathrm{d}u } $

Find the solution to the integral equation using Laplace Transforms.

This requires the convolution theorem:

$\displaystyle \int_0^t{f\left( u \right) \,g\left( t- u \right) \,\mathrm{d}u } = F\left( s \right) \,G\left( s \right) $

In this case, $\displaystyle g\left( t - u \right) = \mathrm{e}^{-3\,\left( t - u \right) } \implies g\left( t \right) = \mathrm{e}^{-3\,t } \implies G\left( s \right) = \frac{1}{s + 3}$.

So upon taking the Laplace Transform of the integral equation, we have

$\displaystyle \begin{align*} F\left( s \right) &= \frac{7}{s^2} - 3\,F\left( s \right) \left( \frac{1}{s + 3} \right) \\
F\left( s \right) &= \frac{7}{s^2} - \frac{3\,F\left( s \right) }{s + 3} \\
F\left( s \right) + \frac{3\,F\left( s \right) }{s + 3} &= \frac{7}{s^2} \\
\left( 1 + \frac{3}{s + 3} \right) F\left( s \right) &= \frac{7}{s^2} \\
\left( \frac{s + 6}{s + 3} \right) F\left( s \right) &= \frac{7}{s^2} \\
F\left( s \right) &= \frac{7 \left( s + 3 \right) }{s^2\,\left( s + 6 \right) } \\
F\left( s\right) &= \frac{7\,s + 21}{s^2\,\left( s + 6 \right) } \end{align*}$

Taking the Inverse Transform will require Partial Fractions:

$\displaystyle \begin{align*} \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + 6} &\equiv \frac{7\,s + 21}{s^2\,\left( s + 6 \right) } \\
A\,s\left( s + 6 \right) + B\,\left( s + 6 \right) + C\,s^2 &= 7\,s + 21 \end{align*}$

Let $\displaystyle s = 0 \implies 6\,B = 21 \implies B = \frac{7}{2} $

Let $\displaystyle s = -6 \implies 36\,C = -21 \implies C = -\frac{7}{12} $

Thus $\displaystyle A\,s\left( s + 6 \right) + \frac{7}{2} \left( s + 6 \right) - \frac{7}{12}\,s^2 = 7\,s + 21 $.

Let $\displaystyle s = 1 $

$\displaystyle \begin{align*} 7\,A + \frac{7}{2} \cdot 7 - \frac{7}{12} \cdot 1^2 &= 7\cdot 7 + 21 \\
7\,A + \frac{49}{2} - \frac{7}{12} &= 70 \\
7\,A + \frac{294}{12} - \frac{7}{12} &= \frac{840}{12} \\
7\,A + \frac{287}{12} &= \frac{840}{12} \\
7\,A &= \frac{553}{12} \\
A &= \frac{79}{12} \end{align*}$

$\displaystyle \begin{align*} F\left( s \right) &= \frac{79}{12} \left( \frac{1}{s} \right) + \frac{7}{2} \left( \frac{1}{s^2} \right) - \frac{7}{12} \left( \frac{1}{s + 6} \right) \\
f\left( t \right) &= \frac{79}{12} + \frac{7}{2}\,t - \frac{7}{12} \,\mathrm{e}^{-6\,t} \end{align*}$

 

[benorin]

This is your work:

This requires the convolution theorem:

$\displaystyle \int_0^t{f\left( u \right) \,g\left( t- u \right) \,\mathrm{d}u } = F\left( s \right) \,G\left( s \right) $

In this case, $\displaystyle g\left( t - u \right) = \mathrm{e}^{-3\,\left( t - u \right) } \implies g\left( t \right) = \mathrm{e}^{-3\,t } \implies G\left( s \right) = \frac{1}{s + 3}$.

So upon taking the Laplace Transform of the integral equation, we have

Edit starts here: (look for the boxes, the first box is an extra 3, all the other boxes are the corrections for removing said 3).

$\displaystyle \begin{align*} F\left( s \right) &= \frac{7}{s^2} - 3\,F\left( s \right) \left( \frac{1}{s + 3} \right) \\
F\left( s \right) &= \frac{7}{s^2} - \frac{\underbrace{\boxed{3}}_{\text{this is the extra 3 I removed from here on out}}\, F\left( s \right) }{s + 3} \\
F\left( s \right) + \frac{F\left( s \right) }{s + 3} &= \frac{7}{s^2} \\
\left( 1 + \frac{\boxed{1}}{s + 3} \right) F\left( s \right) &= \frac{7}{s^2} \\
\left( \frac{s +\boxed{4}}{s + 3} \right) F\left( s \right) &= \frac{7}{s^2} \\
F\left( s \right) &= \frac{7 \left( s + 3 \right) }{s^2\,\left( s +\boxed{4} \right) } \\
F\left( s\right) &= \frac{7\,s + 21}{s^2\,\left( s +\boxed{4}\right) } \end{align*}$

Taking the Inverse Transform will require Partial Fractions:

$\displaystyle \begin{align*} \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s + \boxed{4}} &\equiv \frac{7\,s + 21}{s^2\,\left( s + \boxed{4}\right) } \\
A\,s\left( s +\boxed{4}\right) + B\,\left( s + \boxed{4}\right) + C\,s^2 &= 7\,s + 21 \end{align*}$

Let $\displaystyle s = 0 \implies\boxed{4}\,B = 21 \implies B = \frac{21}{\boxed{4}} $

Let $\displaystyle s = -4 \implies\boxed{16}\,C =\boxed{-7} \implies C =\boxed{ -\frac{7}{16}} $

Thus $\displaystyle A\,s\left( s + \boxed{4}\right) +\boxed{-\frac{7}{16}} \left( s + \boxed{4}\right) +\boxed{- \frac{7}{16}}\,s^2 = 7\,s + 21 $.

Let $\displaystyle s = 1 $

I’m going to stop here.