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Muhammed's question via email about solving a DE using Fourier Transforms

Prove It

Well-known member
MHB Math Helper
Jan 26, 2012
1,403
a) Find the Inverse Fourier Transform of $\displaystyle \begin{align*} \frac{1}{ \omega ^2 - 7\mathrm{i} \, \omega - 10 } \end{align*}$.

b) Hence find the Inverse Fourier Transform of $\displaystyle \begin{align*} \frac{\mathrm{e}^{-3\mathrm{i}\,\omega}}{\omega ^2 - 7\mathrm{i}\,\omega - 10} \end{align*}$.

c) Using these results, solve the following Differential Equation:

$\displaystyle \begin{align*} \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 7\,\frac{\mathrm{d}y}{\mathrm{d}t} + 10y = \delta \left( t - 3 \right) \end{align*}$
a) Start by writing $\displaystyle \begin{align*} \frac{ 1}{ \omega ^2 - 7\mathrm{i}\,\omega - 10 } = \frac{1}{ \omega ^2 - 7\mathrm{i} \,\omega + 10\mathrm{i}^2 } = \frac{1}{ \left( \omega - 5\mathrm{i} \right) \left( \omega - 2\mathrm{i} \right) } \end{align*}$

Now applying partial fractions:

$\displaystyle \begin{align*} \frac{A}{\omega - 5\mathrm{i}} + \frac{B}{ \omega - 2\mathrm{i} } &\equiv \frac{1}{ \left( \omega - 5\mathrm{i} \right) \left( \omega - 2\mathrm{i} \right) } \\ A \left( \omega - 2\mathrm{i} \right) + B \left( \omega - 5\mathrm{i} \right) &\equiv 1 \end{align*}$

Let $\displaystyle \begin{align*} \omega = 5\mathrm{i} \end{align*}$ to find $\displaystyle \begin{align*} 3\mathrm{i}\,A = 1 \implies A = -\frac{\mathrm{i}}{3} \end{align*}$ and let $\displaystyle \begin{align*} \omega = 2\mathrm{i} \end{align*}$ to find $\displaystyle \begin{align*} -3\mathrm{i}\,B = 1 \implies B = \frac{\mathrm{i}}{3} \end{align*}$, giving

$\displaystyle \begin{align*} -\frac{\mathrm{i}}{3} \left( \frac{1}{\omega - 5\mathrm{i}} \right) + \frac{\mathrm{i}}{3} \left( \frac{1}{\omega - 2\mathrm{i}} \right) \equiv \frac{1}{ \left( \omega - 5\mathrm{i} \right) \left( \omega - 2\mathrm{i} \right) } \end{align*}$

Now some rearrangement to apply the rule $\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{1}{ a + \mathrm{i}\,\omega} \right\} = \mathrm{e}^{-a\,t}\,\mathrm{H}(t) \end{align*}$...

$\displaystyle \begin{align*} -\frac{\mathrm{i}}{3} \left( \frac{1}{\omega - 5\mathrm{i}} \right) + \frac{\mathrm{i}}{3} \left( \frac{1}{\omega - 2\mathrm{i}} \right) &= -\frac{\mathrm{i}}{3} \left( \frac{1}{\omega - 5\mathrm{i}} \right) \frac{\mathrm{i}}{\mathrm{i}} + \frac{\mathrm{i}}{3} \left( \frac{1}{\omega - 2\mathrm{i}} \right) \frac{\mathrm{i}}{\mathrm{i}} \\ &= \frac{1}{3} \left( \frac{1}{5 + \mathrm{i}\,\omega} \right) - \frac{1}{3} \left( \frac{1}{2 + \mathrm{i}\,\omega} \right) \end{align*}$

and so

$\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{1}{\omega ^2 - 7\mathrm{i}\,\omega - 10 } \right\} &= \mathcal{ F}^{-1} \left\{ \frac{1}{3} \left( \frac{1}{5 + \mathrm{i}\,\omega } \right) - \frac{1}{3} \left( \frac{1}{2 + \mathrm{i}\,\omega } \right) \right\} \\ &= \frac{1}{3}\mathrm{e}^{-5t}\,\mathrm{H}(t) - \frac{1}{3}\mathrm{e}^{-2t}\,\mathrm{H}(t) \end{align*}$



b) Applying the second shift theorem: $\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \mathrm{e}^{-\mathrm{i}\,\omega\,u } F \left( \omega \right) \right\} = f \left( t - u \right) \end{align*}$, we can see that

$\displaystyle \begin{align*} \mathcal{F}^{-1} \left\{ \frac{ \mathrm{e}^{-3\mathrm{i}\,\omega} }{ \omega ^2 - 7\mathrm{i} \,\omega - 10 } \right\} &= \mathcal{F}^{-1} \left\{ \mathrm{e}^{-3\mathrm{i}\,\omega } \left( \frac{1}{ \omega ^2 - 7\mathrm{i}\,\omega - 10 } \right) \right\} \\ &= \frac{1}{3} \mathrm{e}^{-5 \left( t - 3 \right) }\,\mathrm{H} \left( t - 3 \right) - \frac{1}{3} \mathrm{e}^{-2 \left( t - 3 \right) } \, \mathrm{H} \left( t - 3 \right) \end{align*}$



c) Here we need to use the Operational Theorem $\displaystyle \begin{align*} \mathcal{ F} \left\{ \frac{\mathrm{d}^n f}{\mathrm{d}t^n} \right\} = \left( \mathrm{i}\,\omega \right) ^n F \left( \omega \right) \end{align*}$ and the rule $\displaystyle \begin{align*} \mathcal{F} \left\{ \delta \left( t - T \right) \right\} = \mathrm{e}^{-\mathrm{i}\,T\,\omega } \end{align*}$. Taking the Fourier Transform of both sides gives:

$\displaystyle \begin{align*} \mathcal{F} \left\{ \frac{\mathrm{d}^2y}{\mathrm{d}t^2} + 7\,\frac{\mathrm{d}y}{\mathrm{d}t} + 10y \right\} &= \mathcal{F} \left\{ \delta \left( t - 3 \right) \right\} \\ \left( \mathrm{i}\,\omega \right) ^2 \, Y \left( \omega \right) + 7 \left( \mathrm{i} \,\omega \right) Y \left( \omega \right) + 10 \, Y \left( \omega \right) &= \mathrm{e}^{-3\mathrm{i}\,\omega } \\ -\omega ^2 \, Y \left( \omega \right) + 7\mathrm{i}\,\omega \, Y \left( \omega \right) + 10 \, Y \left( \omega \right) &= \mathrm{e}^{-3\mathrm{i}\,\omega } \\ \left( -\omega ^2 + 7\mathrm{i}\,\omega + 10 \right) Y \left( \omega \right) &= \mathrm{e}^{-3\mathrm{i}\,\omega } \\ - \left( \omega ^2 - 7\mathrm{i}\,\omega - 10 \right) Y\left( \omega \right) &= \mathrm{e}^{-3\mathrm{i}\,\omega } \\ Y \left( \omega \right) &= - \frac{\mathrm{e}^{-3\mathrm{i}\,\omega }}{\omega ^2 - 7\mathrm{i}\,\omega - 10 } \\ y(t) &= \mathcal{F}^{-1} \left\{ -\frac{\mathrm{e}^{-3\mathrm{i}\,\omega}}{\omega ^2 - 7\mathrm{i}\,\omega - 10 } \right\} \\ &= -\frac{1}{3}\mathrm{e}^{-5 \left( t - 3 \right) }\,\mathrm{H} \left( t - 3 \right) + \frac{1}{3}\mathrm{e}^{-2 \left( t - 3 \right) }\,\mathrm{H} \left( t - 3 \right) \end{align*}$