2.11 HW9

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2.11.1 Section 6.6 problem 1

Question: Establish

1.

\(f\circledast g=g\circledast f\)

2.

\(f\circledast \left ( g_{1}+g_{2}\right ) =f\circledast g_{1}+f\circledast g_{2}\)

3.

\(f\circledast \left ( g\circledast h\right ) =\left ( f\circledast g\right ) \circledast h\)

2.11.1.1 Part (a)

From definition\[ f\left ( t\right ) \circledast g\left ( t\right ) =\int _{-\infty }^{\infty }f\left ( t-\tau \right ) g\left ( \tau \right ) d\tau \] Let \(u=t-\tau \), hence \(\frac{du}{d\tau }=-1\). When \(\tau =-\infty \rightarrow u=+\infty \) and when \(\tau =+\infty \rightarrow u=-\infty \), hence the above becomes\[ f\left ( t\right ) \circledast g\left ( t\right ) =\int _{+\infty }^{-\infty }f\left ( u\right ) g\left ( t-u\right ) \left ( -du\right ) \] Pulling the minus sign outside and changing the integration limits \[ f\left ( t\right ) \circledast g\left ( t\right ) =\int _{-\infty }^{\infty }g\left ( t-u\right ) f\left ( u\right ) du \] But since \(u\) is arbitrary, we can relabel \(u\) as \(\tau \) in the above. Hence the above RHS can be written as\[ f\left ( t\right ) \circledast g\left ( t\right ) =\int _{-\infty }^{\infty }g\left ( t-\tau \right ) f\left ( \tau \right ) d\tau \] But \(\int _{-\infty }^{\infty }g\left ( t-\tau \right ) f\left ( \tau \right ) d\tau =g\left ( t\right ) \circledast f\left ( t\right ) \), hence\[ f\left ( t\right ) \circledast g\left ( t\right ) =g\left ( t\right ) \circledast f\left ( t\right ) \] QED.

2.11.1.2 Part (b)

From definition\[ f\left ( t\right ) \circledast \left ( g_{1}\left ( t\right ) +g_{2}\left ( t\right ) \right ) =\int _{-\infty }^{\infty }f\left ( t-\tau \right ) \left ( g_{1}\left ( \tau \right ) +g_{1}\left ( \tau \right ) \right ) d\tau \] By linearity of the integral operation, we can break the integral above\[ \int _{-\infty }^{\infty }f\left ( t-\tau \right ) \left ( g_{1}\left ( \tau \right ) +g_{1}\left ( \tau \right ) \right ) d\tau =\int _{-\infty }^{\infty }f\left ( t-\tau \right ) g_{1}\left ( \tau \right ) d\tau +\int _{-\infty }^{\infty }f\left ( t-\tau \right ) g_{1}\left ( \tau \right ) d\tau \] But \(\int _{-\infty }^{\infty }f\left ( t-\tau \right ) g_{1}\left ( \tau \right ) d\tau =f\left ( t\right ) \circledast g_{1}\left ( t\right ) \) and \(\int _{-\infty }^{\infty }f\left ( t-\tau \right ) g_{1}\left ( \tau \right ) d\tau =f\left ( t\right ) \circledast g_{2}\left ( t\right ) \), hence the above becomes\[ \int _{-\infty }^{\infty }f\left ( t-\tau \right ) \left ( g_{1}\left ( \tau \right ) +g_{1}\left ( \tau \right ) \right ) d\tau =\left ( f\left ( t\right ) \circledast g_{1}\left ( t\right ) \right ) +\left ( f\left ( t\right ) \circledast g_{2}\left ( t\right ) \right ) \] Therefore\[ f\left ( t\right ) \circledast \left ( g_{1}\left ( t\right ) +g_{2}\left ( t\right ) \right ) =\left ( f\left ( t\right ) \circledast g_{1}\left ( t\right ) \right ) +\left ( f\left ( t\right ) \circledast g_{2}\left ( t\right ) \right ) \] QED.

2.11.1.3 Part (c)

From definition\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }\left ( f\circledast g\right ) \left ( \tau \right ) h\left ( t-\tau \right ) d\tau \\ & =\int _{\Re }\left [ \int _{\Re }f\left ( \tau _{1}\right ) g\left ( \tau -\tau _{1}\right ) d\tau _{1}\right ] h\left ( t-\tau \right ) d\tau \\ & =\int _{\Re }\int _{\Re }f\left ( \tau _{1}\right ) g\left ( \tau -\tau _{1}\right ) h\left ( t-\tau \right ) d\tau _{1}d\tau \end{align*}

By Fubini, we can change order of integration\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }\int _{\Re }f\left ( \tau _{1}\right ) g\left ( \tau -\tau _{1}\right ) h\left ( t-\tau \right ) d\tau d\tau _{1}\\ & =\int _{\Re }f\left ( \tau _{1}\right ) \left [ \int _{\Re }g\left ( \tau -\tau _{1}\right ) h\left ( t-\tau \right ) d\tau \right ] d\tau _{1} \end{align*}

By translation, if we add \(\tau _{1}\) to \(\tau \) for both functions in the inner integral above, we obtain\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }f\left ( \tau _{1}\right ) \left [ \int _{\Re }g\left ( \left ( \tau +\tau _{1}\right ) -\tau _{1}\right ) h\left ( t-\left ( \tau +\tau _{1}\right ) \right ) d\tau \right ] d\tau _{1}\\ & =\int _{\Re }f\left ( \tau _{1}\right ) \left [ \int _{\Re }g\left ( \tau \right ) h\left ( \left ( t-\tau _{1}\right ) -\tau \right ) d\tau \right ] d\tau _{1} \end{align*}

But now we see that inner integral is \(\int _{\Re }g\left ( \tau \right ) h\left ( \left ( t-\tau _{1}\right ) -\tau \right ) d\tau =\left ( g\circledast h\right ) \left ( t-\tau _{1}\right ) \), hence the above becomes\begin{align*} \left ( \left ( f\circledast g\right ) \circledast h\right ) \left ( t\right ) & =\int _{\Re }f\left ( \tau _{1}\right ) \left ( g\circledast h\right ) \left ( t-\tau _{1}\right ) d\tau _{1}\\ & =\left ( f\circledast \left ( g\circledast h\right ) \right ) \left ( t\right ) \end{align*}

QED

2.11.2 Section 6.6 problem 2

Find an example showing \(\left ( f\circledast 1\right ) \left ( t\right ) \) need not be equal to \(f\left ( t\right ) \)

Solution Let \(f\left ( t\right ) =e^{t}\), hence\begin{align*} \left ( f\circledast 1\right ) \left ( t\right ) & =\int _{0}^{t}f\left ( t-\tau \right ) \times 1d\tau \\ & =\int _{0}^{t}e^{\left ( t-\tau \right ) }d\tau \\ & =\left [ \frac{e^{\left ( t-\tau \right ) }}{-1}\right ] _{\tau =0}^{\tau =t}\\ & =-\left [ e^{\left ( t-t\right ) }-e^{\left ( t-0\right ) }\right ] \\ & =-\left [ e^{0}-e^{t}\right ] \\ & =-\left ( 1-e^{t}\right ) \\ & =e^{t}-1 \end{align*}

Which is not the same as \(e^{t}\).  QED

2.11.3 Section 6.6 problem 3

Show that \(\left ( f\circledast f\right ) \left ( t\right ) \) is not necessarily non-negative, using \(f\left ( t\right ) =\sin \left ( t\right ) \)

Solution From definition\[ \left ( f\circledast f\right ) \left ( t\right ) =\int _{0}^{t}\sin \left ( \tau \right ) \sin \left ( t-\tau \right ) d\tau \] Using \(\sin A\sin B=\frac{1}{2}\left ( \cos \left ( A-B\right ) -\cos \left ( A+B\right ) \right ) \) on the integrand gives\begin{align*} \left ( f\circledast f\right ) \left ( t\right ) & =\int _{0}^{t}\frac{1}{2}\left ( \cos \left ( \tau -\left ( t-\tau \right ) \right ) -\cos \left ( \tau +\left ( t-\tau \right ) \right ) \right ) d\tau \\ & =\frac{1}{2}\int _{0}^{t}\cos \left ( \tau -\left ( t-\tau \right ) \right ) d\tau -\frac{1}{2}\int _{0}^{t}\cos \left ( t\right ) d\tau \\ & =\frac{1}{2}\int _{0}^{t}\cos \left ( 2\tau -t\right ) d\tau -\frac{1}{2}\int _{0}^{t}\cos \left ( t\right ) d\tau \end{align*}

For the second integral above, since it is w.r.t \(\tau \), then we can pull \(\cos \left ( t\right ) \) outside, which gives\begin{align*} \left ( f\circledast f\right ) \left ( t\right ) & =\frac{1}{2}\left ( \frac{\sin \left ( 2\tau -t\right ) }{2}\right ) _{\tau =0}^{\tau =t}-\frac{1}{2}\cos \left ( t\right ) \int _{0}^{t}d\tau \\ & =\frac{1}{4}\left ( \sin \left ( 2t-t\right ) -\sin \left ( -t\right ) \right ) -\frac{1}{2}t\cos t\\ & =\frac{1}{4}\left ( \sin \left ( t\right ) +\sin \left ( t\right ) \right ) -\frac{1}{2}t\cos t\\ & =\frac{1}{2}\sin t-\frac{1}{2}t\cos t \end{align*}

Let \(t=2\pi \) then\begin{align*} \left ( f\circledast f\right ) \left ( t\right ) & =0-\frac{1}{2}\left ( 2\pi \right ) \\ & =-\pi \end{align*}

Which is negative. Hence we showed that \(\left ( f\circledast f\right ) \left ( t\right ) \) can be negative at some \(t\). QED.

2.11.4 Section 6.6 problem 4

Find Laplace transform of \(f\left ( t\right ) =\int _{0}^{t}\left ( t-\tau \right ) ^{2}\cos \left ( 2\tau \right ) d\tau \)

Solution We see that \[ f\left ( t\right ) =t^{2}\circledast \cos \left ( 2t\right ) \] Therefore, using convolution theorem \[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\mathcal{L}\left \{ t^{2}\right \} \mathcal{L}\left \{ \cos \left ( 2t\right ) \right \} \] But \(\mathcal{L}\left \{ t^{2}\right \} =\frac{2}{s^{3}}\) and \(\mathcal{L}\left \{ \cos \left ( 2t\right ) \right \} =\frac{s}{s^{2}+4}\), hence the above becomes\begin{align*} \mathcal{L}\left \{ f\left ( t\right ) \right \} & =\left ( \frac{2}{s^{3}}\right ) \left ( \frac{s}{s^{2}+4}\right ) \\ & =\frac{2}{s^{2}}\frac{1}{s^{2}+4} \end{align*}

2.11.5 Section 6.6 problem 5

Find Laplace transform of \(f\left ( t\right ) =\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin \left ( \tau \right ) d\tau \)

Solution We see that \[ f\left ( t\right ) =e^{-t}\circledast \sin \left ( t\right ) \] Therefore, using convolution theorem\[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\mathcal{L}\left \{ e^{-t}\right \} \mathcal{L}\left \{ \sin \left ( t\right ) \right \} \] But \(\mathcal{L}\left \{ e^{-t}\right \} =\frac{1}{s+1}\) and \(\mathcal{L}\left \{ \sin \left ( t\right ) \right \} =\frac{1}{s^{2}+1}\), hence the above becomes\[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\frac{1}{\left ( s+1\right ) \left ( s^{2}+1\right ) }\]

2.11.6 Section 6.6 problem 6

Find Laplace transform of \(f\left ( t\right ) =\int _{0}^{t}\left ( t-\tau \right ) e^{\tau }d\tau \)

Solution We see that \[ f\left ( t\right ) =t\circledast e^{t}\] Therefore, using convolution theorem \[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\mathcal{L}\left \{ t\right \} \mathcal{L}\left \{ e^{t}\right \} \] But \(\mathcal{L}\left \{ t\right \} =\frac{1}{s^{2}}\) and \(\mathcal{L}\left \{ e^{t}\right \} =\frac{1}{s-1}\), hence the above becomes\[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\left ( \frac{1}{s^{2}}\right ) \left ( \frac{1}{s-1}\right ) \]

2.11.7 Section 6.6 problem 7

Find Laplace transform of \(f\left ( t\right ) =\int _{0}^{t}\sin \left ( t-\tau \right ) \cos \tau d\tau \)

Solution We see that \[ f\left ( t\right ) =\sin \left ( t\right ) \circledast \cos \left ( t\right ) \] Therefore, using convolution theorem \[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\mathcal{L}\left \{ \sin t\right \} \mathcal{L}\left \{ \cos t\right \} \] But \(\mathcal{L}\left \{ \sin t\right \} =\frac{1}{s^{2}+1}\) and \(\mathcal{L}\left \{ \cos t\right \} =\frac{s}{s^{2}+1}\), hence the above becomes\[\mathcal{L}\left \{ f\left ( t\right ) \right \} =\left ( \frac{1}{s^{2}+1}\right ) \left ( \frac{s}{s^{2}+1}\right ) \]

2.11.8 Section 6.6 problem 8

Find the inverse Laplace transform of \(F\left ( s\right ) =\frac{1}{s^{4}\left ( s^{2}+1\right ) }\) using convolution theorem.

Solution We see that \begin{align*} F\left ( s\right ) & =\frac{1}{s^{4}}\frac{1}{s^{2}+1}\\ & =\mathcal{L}\left ( \frac{t^{3}}{6}\right ) \mathcal{L}\left ( \sin t\right ) \end{align*}

Hence, using convolution theorem\begin{align*} f\left ( t\right ) & =\frac{t^{3}}{6}\circledast \sin t\\ & =\frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau \end{align*}

Integrate by parts. \(\int udv=uv-\int vdu\). Let \(u=\left ( t-\tau \right ) ^{3},dv=\sin \tau \rightarrow du=-3\left ( t-\tau \right ) ^{2},v=-\cos \tau \), hence\begin{align*} \frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau & =\frac{1}{6}\left ( -\left [ \left ( t-\tau \right ) ^{3}\cos \tau \right ] _{0}^{t}-\int _{0}^{t}-3\left ( t-\tau \right ) ^{2}\left ( -\cos \tau \right ) \ d\tau \right ) \\ & =\frac{1}{6}\left ( -\left [ \left ( t-t\right ) ^{3}\cos t-\left ( t-0\right ) ^{3}\cos 0\right ] -3\int _{0}^{t}\left ( t-\tau \right ) ^{2}\left ( \cos \tau \right ) \ d\tau \right ) \\ & =\frac{1}{6}\left ( -\left [ 0-t^{3}\right ] -3\int _{0}^{t}\left ( t-\tau \right ) ^{2}\left ( \cos \tau \right ) \ d\tau \right ) \\ & =\frac{1}{6}\left ( t^{3}-3\int _{0}^{t}\left ( t-\tau \right ) ^{2}\left ( \cos \tau \right ) \ d\tau \right ) \end{align*}

Integrate by parts. Let \(u=\left ( t-\tau \right ) ^{2},dv=\cos \tau \rightarrow du=-2\left ( t-\tau \right ) ,v=\sin \tau \), hence\begin{align*} \frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau & =\frac{1}{6}\left ( t^{3}-3\left [ \left ( \left ( t-\tau \right ) ^{2}\sin \tau \right ) _{0}^{t}-\int _{0}^{t}-2\left ( t-\tau \right ) \sin \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-3\left [ \left ( \left ( t-t\right ) ^{2}\sin t-\left ( t-0\right ) ^{2}\sin 0\right ) _{0}^{t}+2\int _{0}^{t}\left ( t-\tau \right ) \sin \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-3\left [ 0+2\int _{0}^{t}\left ( t-\tau \right ) \sin \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\int _{0}^{t}\left ( t-\tau \right ) \sin \tau d\tau \right ) \end{align*}

Integrate by parts. Let \(u=\left ( t-\tau \right ) ,dv=\sin \tau \rightarrow du=-1,v=-\cos \tau \), hence above becomes\begin{align*} \frac{1}{6}\int _{0}^{t}\left ( t-\tau \right ) ^{3}\sin \tau \ d\tau & =\frac{1}{6}\left ( t^{3}-6\left [ \left ( -\left ( t-\tau \right ) \cos \tau \right ) _{0}^{t}-\int _{0}^{t}\cos \tau d\tau \right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\left [ -\left ( \left ( t-t\right ) \cos t-\left ( t-0\right ) \cos 0\right ) -\left ( \sin \tau \right ) _{0}^{t}\right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\left [ -\left ( 0-t\right ) -\sin t\right ] \right ) \\ & =\frac{1}{6}\left ( t^{3}-6\left ( t-\sin t\right ) \right ) \\ & =\frac{1}{6}\left ( t^{3}-6t+6\sin t\right ) \end{align*}

Hence\[ f\left ( t\right ) =\frac{1}{6}\left ( t^{3}-6t+6\sin t\right ) \]

2.11.9 Section 6.6 problem 9

Find the inverse Laplace transform of \(F\left ( s\right ) =\frac{s}{\left ( s+1\right ) \left ( s^{2}+4\right ) }\) using convolution theorem.

Solution We see that \begin{align*} F\left ( s\right ) & =\frac{1}{s+1}\frac{s}{s^{2}+4}\\ & =\mathcal{L}\left ( e^{-t}\right ) \mathcal{L}\left ( \cos 2t\right ) \end{align*}

Hence, using convolution theorem\begin{align*} f\left ( t\right ) & =e^{-t}\circledast \cos 2t\\ & =\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau \end{align*}

Integrate by parts. \(\int udv=uv-\int vdu\). Let \(u=\cos 2\tau ,dv=e^{-\left ( t-\tau \right ) }\rightarrow du=-2\sin 2\tau ,v=e^{-\left ( t-\tau \right ) }\), hence\begin{align*} \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau & =\left ( \cos 2\tau e^{-\left ( t-\tau \right ) }\right ) _{0}^{t}-\int _{0}^{t}e^{-\left ( t-\tau \right ) }\left ( -2\sin 2\tau \right ) d\tau \\ & =\left ( \cos 2te^{-\left ( t-t\right ) }-\cos 0e^{-\left ( t-0\right ) }\right ) +2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau d\tau \\ & =\left ( \cos 2t-e^{-t}\right ) +2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau d\tau \end{align*}

Integrate by parts. Let \(u=\sin 2\tau ,dv=e^{-\left ( t-\tau \right ) }\rightarrow du=2\cos 2\tau ,v=e^{-\left ( t-\tau \right ) }\), hence\begin{align*} \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau & =\left ( \cos 2t-e^{-t}\right ) +2\left [ \left ( \sin 2\tau e^{-\left ( t-\tau \right ) }\right ) _{0}^{t}-\int _{0}^{t}e^{-\left ( t-\tau \right ) }2\cos 2\tau d\tau \right ] \\ & =\left ( \cos 2t-e^{-t}\right ) +2\left [ \left ( \sin 2te^{-\left ( t-t\right ) }-0\right ) -2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau \right ] \\ & =\left ( \cos 2t-e^{-t}\right ) +2\left [ \sin 2t-2\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau \right ] \\ & =\left ( \cos 2t-e^{-t}\right ) +2\sin 2t-4\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau \end{align*}

Hence\begin{align*} \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau \ d\tau +4\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau & =\cos 2t-e^{-t}+2\sin 2t\\ 5\int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau & =\cos 2t-e^{-t}+2\sin 2t\\ \int _{0}^{t}e^{-\left ( t-\tau \right ) }\cos 2\tau d\tau & =\frac{1}{5}\left ( \cos 2t-e^{-t}+2\sin 2t\right ) \end{align*}

Therefore\[ f\left ( t\right ) =\frac{1}{5}\left ( \cos 2t-e^{-t}+2\sin 2t\right ) \]

2.11.10 Section 6.6 problem 10

Find the inverse Laplace transform of \(F\left ( s\right ) =\frac{1}{\left ( s+1\right ) ^{2}\left ( s^{2}+4\right ) }\) using convolution theorem.

Solution We see that \begin{align*} F\left ( s\right ) & =\frac{1}{\left ( s+1\right ) ^{2}}\frac{1}{s^{2}+4}\\ & =\mathcal{L}\left ( te^{-t}\right ) \mathcal{L}\left ( \frac{1}{2}\sin 2t\right ) \end{align*}

Hence, using convolution theorem \begin{align*} f\left ( t\right ) & =te^{-t}\circledast \frac{1}{2}\sin 2t\\ & =\frac{1}{2}\int _{0}^{t}\left ( t-\tau \right ) e^{-\left ( t-\tau \right ) }\sin 2\tau \\ & =\frac{1}{2}\int _{0}^{t}te^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau -\frac{1}{2}\int _{0}^{t}\tau e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau \end{align*}

The first integral is \[ \int _{0}^{t}te^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau =t\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau \] This is similar to one we did in problem 10 but now we have \(\sin 2\tau \). Using integration by parts again as before gives\begin{align*} t\int _{0}^{t}e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau & =t\left ( \frac{1}{5}\left ( 2e^{-t}-2\cos 2t+\sin 2t\right ) \right ) \\ & =\frac{t}{5}\left ( 2e^{-t}-2\cos 2t+\sin 2t\right ) \end{align*}

Now we need to evaluate the second integral \(\int _{0}^{t}\tau e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau \). This can also be done using integration by part. But I used CAS here, the result is\[ \int _{0}^{t}\tau e^{-\left ( t-\tau \right ) }\sin 2\tau \ d\tau =\frac{1}{25}\left ( -4e^{-t}+\left ( 4-10t\right ) \cos 2t+\left ( 3+5t\right ) \sin 2t\right ) \] Therefore\begin{align*} f\left ( t\right ) & =\frac{1}{2}\frac{t}{5}\left ( 2e^{-t}-2\cos \left ( 2t\right ) +\sin \left ( 2t\right ) \right ) -\frac{1}{2}\frac{1}{25}\left ( -4e^{-t}+\left ( 4-10t\right ) \cos \left ( 2t\right ) +\left ( 3+5t\right ) \sin \left ( 2t\right ) \right ) \\ & =\frac{2}{25}e^{-t}-\frac{2}{25}\cos 2t-\frac{3}{50}\sin 2t+\frac{1}{5}te^{-t}\allowbreak \end{align*}

2.11.11 Section 6.6 problem 11

Find the inverse Laplace transform of \(F\left ( s\right ) =\frac{G\left ( s\right ) }{s^{2}+1}\) using convolution theorem.

Solution We see that \[ F\left ( s\right ) =G\left ( s\right ) \frac{1}{s^{2}+1}=G\left ( s\right ) \mathcal{L}\left ( \sin t\right ) \] Hence, using convolution theorem \[ f\left ( t\right ) =g\left ( t\right ) \circledast \sin t=\int _{0}^{t}\sin \left ( t-\tau \right ) g\left ( \tau \right ) d\tau \] Or\[ f\left ( t\right ) =\int _{0}^{t}g\left ( t-\tau \right ) \sin \left ( \tau \right ) d\tau \]