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

AI Thread Summary
The discussion focuses on solving the integral equation involving the function f(t) using Laplace Transforms. The convolution theorem is applied, leading to the transformation of the integral equation into a solvable algebraic form. After manipulating the equation, the Laplace Transform F(s) is expressed as F(s) = (7(s + 3))/(s^2(s + 6)). The inverse transform is computed using partial fractions, resulting in the final solution f(t) = (79/12) + (7/2)t - (7/12)e^(-6t). This demonstrates the effective application of Laplace Transforms in solving integral equations.
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$\displaystyle f\left( t \right)$ 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 $

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

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