Find Laplace Transform of \(f\left ( t\right ) =\cosh \left ( bt\right ) \)
solution Since \(\cosh \left ( bt\right ) =\frac{e^{bt}+e^{-bt}}{2}\) then\begin{align*} \mathcal{L}\cosh \left ( bt\right ) & =\frac{1}{2}\mathcal{L}\left ( e^{bt}+e^{-bt}\right ) \\ & =\frac{1}{2}\left ( \mathcal{L}e^{bt}+\mathcal{L}e^{-bt}\right ) \end{align*}
But \[\mathcal{L}e^{bt}=\frac{1}{s-b}\]
For \(s>b\) and
\[\mathcal{L}e^{bt}=\frac{1}{s-b}\]
For \(s<b\). Hence
\begin{align*} \mathcal{L}\cosh \left ( bt\right ) & =\frac{1}{2}\left ( \frac{1}{s-b}+\frac{1}{s+b}\right ) \\ & =\frac{s^{2}}{s^{2}-b^{2}} \end{align*}
For \(s>\left \vert b\right \vert \)
Find Laplace Transform of \(f\left ( t\right ) =\sinh \left ( bt\right ) \)
solution Since \(\sinh \left ( bt\right ) =\frac{e^{bt}-e^{-bt}}{2}\) then\begin{align*} \mathcal{L}\sinh \left ( bt\right ) & =\frac{1}{2}\mathcal{L}\left ( e^{bt}-e^{-bt}\right ) \\ & =\frac{1}{2}\left ( \mathcal{L}e^{bt}-\mathcal{L}e^{-bt}\right ) \end{align*}
But, as we found in the last problem \[\mathcal{L}e^{bt}=\frac{1}{s-b}\qquad s>b \] And\[\mathcal{L}e^{-bt}=\frac{1}{s+b}\qquad s<b \] Therefore\begin{align*} \mathcal{L}\sinh \left ( bt\right ) & =\frac{1}{2}\left ( \frac{1}{s-b}-\frac{1}{s+b}\right ) \qquad s>b;s<b\\ & =\frac{b}{s^{2}-b^{2}}\qquad s>\left \vert b\right \vert \end{align*}
Find Laplace Transform of \(f\left ( t\right ) =e^{at}\cosh \left ( bt\right ) \)
solution Using the property that \[ e^{at}f\left ( t\right ) \Longleftrightarrow F\left ( s-a\right ) \] Where \(f\left ( t\right ) =\cosh \left ( bt\right ) \) now. We already found above that \(\cosh \left ( bt\right ) \Longleftrightarrow \frac{s}{s^{2}-b^{2}}\), for \(s>\left \vert b\right \vert \). In other words, \(F\left ( s\right ) =\frac{s}{s^{2}-b^{2}}\), therefore\[ e^{at}\cosh \left ( bt\right ) \Longleftrightarrow \frac{\left ( s-a\right ) }{\left ( s-a\right ) ^{2}-b^{2}}\qquad s-a>\left \vert b\right \vert \]
Find Laplace Transform of \(f\left ( t\right ) =e^{at}\sinh \left ( bt\right ) \)
solution Using the property that \[ e^{at}f\left ( t\right ) \Longleftrightarrow F\left ( s-a\right ) \] Where \(f\left ( t\right ) =\sinh \left ( bt\right ) \) now. We already found above that \(\sinh \left ( bt\right ) \Longleftrightarrow \frac{b}{s^{2}-b^{2}}\), for \(s>\left \vert b\right \vert \). In other words, \(F\left ( s\right ) =\frac{b}{s^{2}-b^{2}}\), therefore\[ e^{at}\sinh \left ( bt\right ) \Longleftrightarrow \frac{b}{\left ( s-a\right ) ^{2}-b^{2}}\qquad s-a>\left \vert b\right \vert \]
Use Laplace transform to solve \(y^{\left ( 4\right ) }-4y^{\prime \prime \prime }+6y^{\prime \prime }-4y^{\prime }+y=0\) for \(y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1,y^{\prime \prime }\left ( 0\right ) =0,y^{\prime \prime \prime }\left ( 0\right ) =1\)
Solution Taking Laplace transform of the ODE gives
\begin{equation} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} -4\mathcal{L}\left \{ y^{\prime \prime \prime }\right \} +6\mathcal{L}\left \{ y^{\prime \prime }\right \} -4\mathcal{L}\left \{ y^{\prime }\right \} +\mathcal{L}\left \{ y\right \} =0 \tag{1} \end{equation} Let \(\mathcal{L}\left \{ y\right \} =Y\left ( s\right ) \) then\begin{align*} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} & =s^{4}Y\left ( s\right ) -s^{3}y\left ( 0\right ) -s^{2}y^{\prime }\left ( 0\right ) -sy^{\prime \prime }\left ( 0\right ) -y^{\prime \prime \prime }\left ( 0\right ) \\ & =s^{4}Y\left ( s\right ) -s^{3}\left ( 0\right ) -s^{2}\left ( 1\right ) -s\left ( 0\right ) -1\\ & =s^{4}Y\left ( s\right ) -s^{2}-1 \end{align*}
And\begin{align*} \mathcal{L}\left \{ y^{\prime \prime \prime }\right \} & =s^{3}Y\left ( s\right ) -s^{2}y\left ( 0\right ) -sy^{\prime }\left ( 0\right ) -y^{\prime \prime }\left ( 0\right ) \\ & =s^{3}Y\left ( s\right ) -s^{2}\left ( 0\right ) -s\left ( 1\right ) -0\\ & =s^{3}Y\left ( s\right ) -s \end{align*}
And\begin{align*} \mathcal{L}\left \{ y^{\prime \prime }\right \} & =s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \\ & =s^{2}Y\left ( s\right ) -s\left ( 0\right ) -1\\ & =s^{2}Y\left ( s\right ) -1 \end{align*}
And\begin{align*} \mathcal{L}\left \{ y^{\prime }\right \} & =sY\left ( s\right ) -y\left ( 0\right ) \\ & =sY\left ( s\right ) \end{align*}
Hence (1) becomes\begin{align*} \left ( s^{4}Y\left ( s\right ) -s^{2}-1\right ) -4\left ( s^{3}Y\left ( s\right ) -s\right ) +6\left ( s^{2}Y\left ( s\right ) -1\right ) -4\left ( sY\left ( s\right ) \right ) +Y\left ( s\right ) & =0\\ Y\left ( s\right ) \left ( s^{4}-4s^{3}+6s^{2}-4s+1\right ) -s^{2}-1+4s-6 & =0 \end{align*}
Therefore\begin{align} Y\left ( s\right ) & =\frac{s^{2}-4s+7}{s^{4}-4s^{3}+6s^{2}-4s+1}\nonumber \\ & =\frac{s^{2}-4s+7}{\left ( s-1\right ) ^{4}}\nonumber \\ & =\frac{s^{2}}{\left ( s-1\right ) ^{4}}-\frac{4s}{\left ( s-1\right ) ^{4}}+\frac{7}{\left ( s-1\right ) ^{4}} \tag{2} \end{align}
But \begin{align*} \frac{s^{2}}{\left ( s-1\right ) ^{4}} & =\frac{\left ( s-1\right ) ^{2}-1+2s}{\left ( s-1\right ) ^{4}}\\ & =\frac{\left ( s-1\right ) ^{2}}{\left ( s-1\right ) ^{4}}-\frac{1}{\left ( s-1\right ) ^{4}}+2\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{4}}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}-\frac{1}{\left ( s-1\right ) ^{4}}+2\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{4}}+2\frac{1}{\left ( s-1\right ) ^{4}}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}-\frac{1}{\left ( s-1\right ) ^{4}}+2\frac{1}{\left ( s-1\right ) ^{3}}+2\frac{1}{\left ( s-1\right ) ^{4}}\\ & =\frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{\left ( s-1\right ) ^{3}}+\frac{1}{\left ( s-1\right ) ^{4}} \end{align*}
And\begin{align*} \frac{4s}{\left ( s-1\right ) ^{4}} & =4\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{4}}\\ & =4\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{4}}+4\frac{1}{\left ( s-1\right ) ^{4}}\\ & =\frac{4}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}} \end{align*}
Therefore (2) becomes\begin{align} Y\left ( s\right ) & =\left ( \frac{1}{\left ( s-1\right ) ^{2}}+\frac{2}{\left ( s-1\right ) ^{3}}+\frac{1}{\left ( s-1\right ) ^{4}}\right ) -\left ( \frac{4}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}}\right ) +\frac{7}{\left ( s-1\right ) ^{4}}\nonumber \\ & =\frac{1}{\left ( s-1\right ) ^{2}}-\frac{2}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}} \tag{3} \end{align}
Now using property the shift property of \(F\left ( s\right ) \) together with\begin{align*} \frac{1}{s^{2}} & \Longleftrightarrow t\\ \frac{1}{s^{3}} & \Longleftrightarrow \frac{t^{2}}{2}\\ \frac{1}{s^{4}} & \Longleftrightarrow \frac{t^{3}}{6} \end{align*}
Therefore\begin{align*} \frac{1}{\left ( s-1\right ) ^{2}} & \Longleftrightarrow e^{t}t\\ \frac{1}{\left ( s-1\right ) ^{3}} & \Longleftrightarrow e^{t}\frac{t^{2}}{2}\\ \frac{1}{\left ( s-1\right ) ^{4}} & \Longleftrightarrow e^{t}\frac{t^{3}}{6} \end{align*}
And (3) becomes\begin{align*} \frac{1}{\left ( s-1\right ) ^{2}}-\frac{2}{\left ( s-1\right ) ^{3}}+\frac{4}{\left ( s-1\right ) ^{4}} & \Longleftrightarrow e^{t}t-2\left ( e^{t}\frac{t^{2}}{2}\right ) +4\left ( e^{t}\frac{t^{3}}{6}\right ) \\ & =e^{t}t-e^{t}t^{2}+\frac{2}{3}e^{t}t^{3} \end{align*}
Hence\[ y\left ( t\right ) =e^{t}\left ( t-t^{2}+\frac{2}{3}t^{3}\right ) \]
Use Laplace transform to solve \(y^{\left ( 4\right ) }-y=0\) for \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0,y^{\prime \prime }\left ( 0\right ) =1,y^{\prime \prime \prime }\left ( 0\right ) =0\)
Solution Taking Laplace transform of the ODE gives\begin{equation} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} -\mathcal{L}\left \{ y\right \} =0 \tag{1} \end{equation} Let \(\mathcal{L}\left \{ y\right \} =Y\left ( s\right ) \) then\begin{align*} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} & =s^{4}Y\left ( s\right ) -s^{3}y\left ( 0\right ) -s^{2}y^{\prime }\left ( 0\right ) -sy^{\prime \prime }\left ( 0\right ) -y^{\prime \prime \prime }\left ( 0\right ) \\ & =s^{4}Y\left ( s\right ) -s^{3}\left ( 1\right ) -s^{2}\left ( 0\right ) -s\left ( 1\right ) -0\\ & =s^{4}Y\left ( s\right ) -s^{3}-s \end{align*}
Hence (1) becomes\[ s^{4}Y\left ( s\right ) -s^{3}-s-Y\left ( s\right ) =0 \] Solving for \(Y\left ( s\right ) \) gives\begin{align*} Y\left ( s\right ) & =\frac{s^{3}+s}{s^{4}-1}\\ & =\frac{s\left ( s^{2}+1\right ) }{s^{4}-1}\\ & =\frac{s\left ( s^{2}+1\right ) }{\left ( s^{2}-1\right ) \left ( s^{2}+1\right ) }\\ & =\frac{s}{s^{2}-1} \end{align*}
But, Hence above becomes, where \(a=1\)\[ \frac{s}{s^{2}-1}\Longleftrightarrow \cosh \left ( t\right ) \] Hence\[ y\left ( t\right ) =\cosh \left ( at\right ) \]
Use Laplace transform to solve \(y^{\left ( 4\right ) }-4y=0\) for \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0,y^{\prime \prime }\left ( 0\right ) =-2,y^{\prime \prime \prime }\left ( 0\right ) =0\)
Solution Taking Laplace transform of the ODE gives\begin{equation} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} -4\mathcal{L}\left \{ y\right \} =0 \tag{1} \end{equation} Let \(\mathcal{L}\left \{ y\right \} =Y\left ( s\right ) \) then\begin{align*} \mathcal{L}\left \{ y^{\left ( 4\right ) }\right \} & =s^{4}Y\left ( s\right ) -s^{3}y\left ( 0\right ) -s^{2}y^{\prime }\left ( 0\right ) -sy^{\prime \prime }\left ( 0\right ) -y^{\prime \prime \prime }\left ( 0\right ) \\ & =s^{4}Y\left ( s\right ) -s^{3}\left ( 1\right ) -s^{2}\left ( 0\right ) -s\left ( -2\right ) -0\\ & =s^{4}Y\left ( s\right ) -s^{3}+2s \end{align*}
Hence (1) becomes\[ s^{4}Y\left ( s\right ) -s^{3}+2s-4Y\left ( s\right ) =0 \] Solving for \(Y\left ( s\right ) \) gives\begin{align*} Y\left ( s\right ) & =\frac{s^{3}-2s}{s^{4}-4}\\ & =\frac{s^{3}-2s}{\left ( s^{2}-2\right ) \left ( s^{2}+2\right ) }\\ & =\frac{s\left ( s^{2}-2\right ) }{\left ( s^{2}-2\right ) \left ( s^{2}+2\right ) }\\ & =\frac{s}{\left ( s^{2}+2\right ) } \end{align*}
Using \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}},\) the above becomes, where \(a=\sqrt{2}\)\[ \frac{s}{\left ( s^{2}+2\right ) }\Longleftrightarrow \cos \left ( \sqrt{2}t\right ) \] Hence\[ y\left ( t\right ) =\cos \left ( \sqrt{2}t\right ) \]
Use Laplace transform to solve \(y^{\prime \prime }+\omega ^{2}y=\cos 2t;\) \(\omega ^{2}\neq 4;\) \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0\)
Solution Let \(Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \} \). Taking Laplace transform of the ODE, and using \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}\) gives\begin{equation} s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) +\omega ^{2}Y\left ( s\right ) =\frac{s}{s^{2}+4} \tag{1} \end{equation} Applying initial conditions\[ s^{2}Y\left ( s\right ) -s+\omega ^{2}Y\left ( s\right ) =\frac{s}{s^{2}+4}\] Solving for \(Y\left ( s\right ) \)\begin{align} Y\left ( s\right ) \left ( s^{2}+\omega ^{2}\right ) -s & =\frac{s}{s^{2}+4}\nonumber \\ Y\left ( s\right ) & =\frac{s}{\left ( s^{2}+4\right ) \left ( s^{2}+\omega ^{2}\right ) }+\frac{s}{\left ( s^{2}+\omega ^{2}\right ) } \tag{2} \end{align}
But \begin{align*} \frac{s}{\left ( s^{2}+4\right ) \left ( s^{2}+\omega ^{2}\right ) } & =\frac{As+B}{\left ( s^{2}+4\right ) }+\frac{Cs+D}{\left ( s^{2}+\omega ^{2}\right ) }\\ s & =\left ( As+B\right ) \left ( s^{2}+\omega ^{2}\right ) +\left ( Cs+D\right ) \left ( s^{2}+4\right ) \\ s & =4D+As^{3}+Bs^{2}+Cs^{3}+B\omega ^{2}+s^{2}D+4Cs+As\omega ^{2}\allowbreak \\ s & =\left ( 4D+B\omega ^{2}\right ) +s\left ( 4C+A\omega ^{2}\right ) +s^{2}\left ( B+D\right ) +s^{3}\left ( A+C\right ) \end{align*}
Hence \begin{align*} 4D+B\omega ^{2} & =0\\ 4C+A\omega ^{2} & =1\\ B+D & =0\\ A+C & =0 \end{align*}
Equation (2,4) gives \(A=\frac{1}{\omega ^{2}-4},C=\frac{1}{4-\omega ^{2}}\) and (1,3) gives \(B=0,D=0\). Hence\[ \frac{s}{\left ( s^{2}+4\right ) \left ( s^{2}+\omega ^{2}\right ) }=\left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{1}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) }\] Therefore (2) becomes\begin{align*} Y\left ( s\right ) & =\left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{1}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) }+\frac{s}{\left ( s^{2}+\omega ^{2}\right ) }\\ & =\left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{5-\omega ^{2}}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) } \end{align*}
Using \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}\), the above becomes\begin{align*} \left ( \frac{1}{\omega ^{2}-4}\right ) \frac{s}{\left ( s^{2}+4\right ) }+\left ( \frac{5-\omega ^{2}}{4-\omega ^{2}}\right ) \frac{s}{\left ( s^{2}+\omega ^{2}\right ) } & \Longleftrightarrow \left ( \frac{1}{\omega ^{2}-4}\right ) \cos \left ( 2t\right ) +\left ( \frac{5-\omega ^{2}}{4-\omega ^{2}}\right ) \cos \left ( \omega t\right ) \\ & =\left ( \frac{1}{\omega ^{2}-4}\right ) \cos \left ( 2t\right ) +\left ( \frac{\omega ^{2}-5}{\omega ^{2}-4}\right ) \cos \left ( \omega t\right ) \end{align*}
Hence\begin{align*} y\left ( t\right ) & =\left ( \frac{1}{\omega ^{2}-4}\right ) \cos \left ( 2t\right ) +\left ( \frac{\omega ^{2}-5}{\omega ^{2}-4}\right ) \cos \left ( \omega t\right ) \\ & =\frac{\left ( \omega ^{2}-5\right ) \cos \left ( \omega t\right ) +\cos \left ( 2t\right ) }{\omega ^{2}-4} \end{align*}
Use Laplace transform to solve \(y^{\prime \prime }-2y^{\prime }+2y=\cos t;\) \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0\)
Solution Let \(Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \} \). Taking Laplace transform of the ODE, and using \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}\) gives\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -2\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +2Y\left ( s\right ) =\frac{s}{s^{2}+1} \tag{1} \end{equation} Applying initial conditions\[ s^{2}Y\left ( s\right ) -s-2\left ( sY\left ( s\right ) -1\right ) +2Y\left ( s\right ) =\frac{s}{s^{2}+1}\] Solving for \(Y\left ( s\right ) \)\begin{align} s^{2}Y\left ( s\right ) -s-2sY\left ( s\right ) +2+2Y\left ( s\right ) & =\frac{s}{s^{2}+1}\nonumber \\ Y\left ( s\right ) \left ( s^{2}-2s+2\right ) -s+2 & =\frac{s}{s^{2}+1}\nonumber \\ Y\left ( s\right ) & =\frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) }+\frac{s}{\left ( s^{2}-2s+2\right ) }-\frac{2}{\left ( s^{2}-2s+2\right ) } \tag{2} \end{align}
But \begin{align*} \frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{As+B}{\left ( s^{2}+1\right ) }+\frac{Cs+D}{s^{2}-2s+2}\\ s & =\left ( As+B\right ) \left ( s^{2}-2s+2\right ) +\left ( Cs+D\right ) \left ( s^{2}+1\right ) \\ s & =2B+D-2As^{2}+As^{3}+Bs^{2}+Cs^{3}+s^{2}D+2As-\allowbreak 2Bs+Cs\\ s & =\left ( 2B+D\right ) +s\left ( 2A-2B+C\right ) +s^{2}\left ( -2A+B+D\right ) +s^{3}\left ( A+C\right ) \end{align*}
Hence\begin{align*} 2B+D & =0\\ 2A-2B+C & =1\\ -2A+B+D & =0\\ A+C & =0 \end{align*}
Solving gives \(A=\frac{1}{5},B=-\frac{2}{5},C=-\frac{1}{5},D=\frac{4}{5}\), hence\begin{align} \frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{s-2}{\left ( s^{2}+1\right ) }-\frac{1}{5}\frac{s-4}{s^{2}-2s+2}\nonumber \\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{s}{s^{2}-2s+2}+\frac{4}{5}\frac{1}{s^{2}-2s+2} \tag{3} \end{align}
Completing the squares for \begin{align*} s^{2}-2s+2 & =a\left ( s+b\right ) ^{2}+d\\ & =a\left ( s^{2}+b^{2}+2bs\right ) +d\\ & =as^{2}+ab^{2}+2abs+d \end{align*}
Hence \(a=1,2ab=-2,\left ( ab^{2}+d\right ) =2\), hence \(b=-1,d=1\), hence\[ s^{2}-2s+2=\left ( s-1\right ) ^{2}+1 \] Hence (3) becomes\begin{align*} \frac{s}{\left ( s^{2}+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{s}{\left ( s-1\right ) ^{2}+1}+\frac{4}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{2}+1}+\frac{4}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}-\frac{1}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{4}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} \end{align*}
Therefore (2) becomes\begin{align*} Y\left ( s\right ) & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{s}{\left ( s-1\right ) ^{2}+1}-\frac{2}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{2}+1}-\frac{2}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{1}{\left ( s-1\right ) ^{2}+1}-\frac{2}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{s}{s^{2}+1}-\frac{2}{5}\frac{1}{s^{2}+1}+\frac{4}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}-\frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} \end{align*}
Using \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}},\sin \left ( at\right ) \Longleftrightarrow \frac{a}{s^{2}+a^{2}}\) and the shift property of Laplace transform, then\begin{align*} \frac{1}{5}\frac{s}{s^{2}+1} & \Longleftrightarrow \frac{1}{5}\cos \left ( t\right ) \\ \frac{2}{5}\frac{1}{s^{2}+1} & \Longleftrightarrow \frac{2}{5}\sin \left ( t\right ) \\ \frac{4}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{4}{5}e^{t}\cos t\\ \frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{8}{5}e^{t}\sin t \end{align*}
Hence\begin{align*} y\left ( t\right ) & =\frac{1}{5}\cos \left ( t\right ) -\frac{2}{5}\sin \left ( t\right ) +\frac{4}{5}e^{t}\cos t-\frac{2}{5}e^{t}\sin t\\ & \frac{1}{5}\left ( \cos t-2\sin t+4e^{t}\cos t-2e^{t}\sin t\right ) \end{align*}
Use Laplace transform to solve \(y^{\prime \prime }-2y^{\prime }+2y=e^{-t};y\left ( 0\right ) =0,y^{\prime }\left ( 0\right ) =1\)
Solution Let \(Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \} \). Taking Laplace transform of the ODE, and using \(e^{-t}\Longleftrightarrow \frac{1}{s+1}\) gives\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -2\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +2Y\left ( s\right ) =\frac{1}{s+1} \tag{1} \end{equation} Applying initial conditions gives\[ s^{2}Y\left ( s\right ) -1-2sY\left ( s\right ) +2Y\left ( s\right ) =\frac{1}{s+1}\] Solving for \(Y\left ( s\right ) \)\begin{align} Y\left ( s\right ) \left ( s^{2}-2s+2\right ) -1 & =\frac{1}{s+1}\nonumber \\ Y\left ( s\right ) & =\frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) }+\frac{1}{s^{2}-2s+2} \tag{2} \end{align}
But \begin{align*} \frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{A}{s+1}+\frac{Bs+C}{s^{2}-2s+2}\\ 1 & =A\left ( s^{2}-2s+2\right ) +\left ( Bs+C\right ) \left ( s+1\right ) \\ 1 & =2A+C+As^{2}+Bs^{2}-2As+Bs+Cs\\ 1 & =\left ( 2A+C\right ) +s\left ( -2A+B+C\right ) +s^{2}\left ( A+B\right ) \end{align*}
Hence\begin{align*} 1 & =2A+C\\ 0 & =-2A+B+C\\ 0 & =A+B \end{align*}
Solving gives \(A=\frac{1}{5},B=-\frac{1}{5},C=\frac{3}{5}\), therefore\begin{align*} \frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{1}{s+1}+\frac{-\frac{1}{5}s+\frac{3}{5}}{s^{2}-2s+2}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{s}{s^{2}-2s+2}++\frac{3}{5}\frac{1}{s^{2}-2s+2} \end{align*}
Completing the square for \(s^{2}-2s+2\) which was done in last problem, gives \(\left ( s-1\right ) ^{2}+1\), hence the above becomes\begin{align*} \frac{1}{\left ( s+1\right ) \left ( s^{2}-2s+2\right ) } & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{s}{\left ( s-1\right ) ^{2}+1}++\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) +1}{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}-\frac{1}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{3}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}\\ & =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} \end{align*}
Therefore (2) becomes\[ Y\left ( s\right ) =\frac{1}{5}\frac{1}{s+1}-\frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1}+\frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1}+\frac{1}{\left ( s-1\right ) ^{2}+1}\] Using \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}},\sin \left ( at\right ) \Longleftrightarrow \frac{a}{s^{2}+a^{2}}\) and the shift property of Laplace transform, then\begin{align*} \frac{1}{5}\frac{1}{s+1} & \Longleftrightarrow \frac{1}{5}e^{-t}\\ \frac{1}{5}\frac{\left ( s-1\right ) }{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{1}{5}e^{t}\cos t\\ \frac{2}{5}\frac{1}{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow \frac{2}{5}e^{t}\sin t\\ \frac{1}{\left ( s-1\right ) ^{2}+1} & \Longleftrightarrow e^{t}\sin t \end{align*}
Hence\begin{align*} y\left ( t\right ) & =\frac{1}{5}e^{-t}-\frac{1}{5}e^{t}\cos t+\frac{2}{5}e^{t}\sin t+e^{t}\sin t\\ & =\frac{1}{5}\left ( e^{-t}-e^{t}\cos t+7e^{t}\sin t\right ) \end{align*}
Use Laplace transform to solve \(y^{\prime \prime }+2y^{\prime }+y=4e^{-t};y\left ( 0\right ) =2,y^{\prime }\left ( 0\right ) =-1\)
Solution Let \(Y\left ( s\right ) =\mathcal{L}\left \{ y\left ( t\right ) \right \} \). Taking Laplace transform of the ODE, and using \(e^{-t}\Longleftrightarrow \frac{1}{s+1}\) gives\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +Y\left ( s\right ) =\frac{4}{s+1} \tag{1} \end{equation} Applying initial conditions gives\[ \left ( s^{2}Y\left ( s\right ) -2s+1\right ) +2\left ( sY\left ( s\right ) -2\right ) +Y\left ( s\right ) =\frac{4}{s+1}\] Solving for \(Y\left ( s\right ) \)\begin{align*} Y\left ( s\right ) \left ( s^{2}+2s+1\right ) -2s+1-4 & =\frac{4}{s+1}\\ Y\left ( s\right ) \left ( s^{2}+2s+1\right ) & =\frac{4}{s+1}+2s-1+4\\ Y\left ( s\right ) & =\frac{4}{\left ( s+1\right ) \left ( s^{2}+2s+1\right ) }+\frac{2s}{\left ( s^{2}+2s+1\right ) }-\frac{1}{\left ( s^{2}+2s+1\right ) }+\frac{4}{\left ( s^{2}+2s+1\right ) } \end{align*}
But \(\left ( s^{2}+2s+1\right ) =\left ( s+1\right ) ^{2}\), hence\begin{equation} Y\left ( s\right ) =\frac{4}{\left ( s+1\right ) ^{3}}+\frac{2s}{\left ( s+1\right ) ^{2}}-\frac{1}{\left ( s+1\right ) ^{2}}+\frac{4}{\left ( s+1\right ) ^{2}} \tag{2} \end{equation} But \begin{align*} \frac{2s}{\left ( s+1\right ) ^{2}} & =2\frac{s+1-1}{\left ( s+1\right ) ^{2}}\\ & =2\frac{\left ( s+1\right ) }{\left ( s+1\right ) ^{2}}-2\frac{1}{\left ( s+1\right ) ^{2}}\\ & =2\frac{1}{s+1}-2\frac{1}{\left ( s+1\right ) ^{2}} \end{align*}
Hence (2) becomes\begin{equation} Y\left ( s\right ) =\frac{4}{\left ( s+1\right ) ^{3}}+2\frac{1}{s+1}-2\frac{1}{\left ( s+1\right ) ^{2}}-\frac{1}{\left ( s+1\right ) ^{2}}+\frac{4}{\left ( s+1\right ) ^{2}} \tag{3} \end{equation} We now ready to do the inversion. Since \(\frac{1}{s^{3}}\Longleftrightarrow \frac{t^{2}}{2}\) and \(\frac{1}{s^{2}}\Longleftrightarrow t\) and \(\frac{1}{s}\Longleftrightarrow 1\) and using the shift property \(e^{at}f\left ( t\right ) \Longleftrightarrow F\left ( s-a\right ) \), then using these into (3) gives\begin{align*} \frac{4}{\left ( s+1\right ) ^{3}} & \Longleftrightarrow 4e^{-t}\left ( \frac{t^{2}}{2}\right ) \\ 2\frac{1}{s+1} & \Longleftrightarrow 2e^{-t}\\ 2\frac{1}{\left ( s+1\right ) ^{2}} & \Longleftrightarrow 2e^{-t}t\\ \frac{1}{\left ( s+1\right ) ^{2}} & \Longleftrightarrow e^{-t}t\\ \frac{4}{\left ( s+1\right ) ^{2}} & \Longleftrightarrow 4e^{-t}t \end{align*}
Now (3) becomes\begin{align*} Y\left ( s\right ) & \Longleftrightarrow 4e^{-t}\left ( \frac{t^{2}}{2}\right ) +2e^{-t}-2e^{-t}t-e^{-t}t+4e^{-t}t\\ & =e^{-t}\left ( 2t^{2}+2-2t-t+4t\right ) \\ & =e^{-t}\left ( 2t^{2}+t+2\right ) \end{align*}
Suppose that \(F\left ( s\right ) =\mathcal{L}\left \{ f\left ( t\right ) \right \} \) exists for \(s>a\geq 0.\)
Solution
From definition, \[\mathcal{L}\left \{ f\left ( ct\right ) \right \} =\int _{0}^{\infty }f\left ( ct\right ) e^{-st}dt \] Let \(ct=\tau \), then when \(t=0,\tau =0\) and when \(t=\infty ,\tau =\infty \), and \(c=\frac{d\tau }{dt}.\) Hence the above becomes\begin{align*} \mathcal{L}\left \{ f\left ( ct\right ) \right \} & =\int _{0}^{\infty }f\left ( \tau \right ) e^{-s\left ( \frac{\tau }{c}\right ) }\frac{d\tau }{c}\\ & =\frac{1}{c}\int _{0}^{\infty }f\left ( \tau \right ) e^{-\tau \left ( \frac{s}{c}\right ) }d\tau \end{align*}
We see from above that \(\mathcal{L}\left \{ f\left ( ct\right ) \right \} \) is \(\frac{1}{c}F\left ( \frac{s}{c}\right ) .\)Now we look at the conditions which makes the above integral converges. Let \[ \left \vert f\left ( \tau \right ) e^{-\tau \left ( \frac{s}{c}\right ) }\right \vert \leq k\left \vert e^{at}e^{-\tau \left ( \frac{s}{c}\right ) }\right \vert \] Where \(k\) is some constant. Then\begin{align*} \int _{0}^{\infty }f\left ( t\right ) e^{-t\left ( \frac{s}{c}\right ) }dt & \leq k\int _{0}^{\infty }e^{at}e^{-t\left ( \frac{s}{c}\right ) }dt\\ & =k\int _{0}^{\infty }e^{-t\left ( \frac{s}{c}-a\right ) }dt \end{align*}
But \(\int _{0}^{\infty }e^{-t\left ( \frac{s}{c}-a\right ) }d\tau \) converges if \(\frac{s}{c}-a>0\) or \[ s>ca \] Hence this is the condition for \(\int _{0}^{\infty }f\left ( t\right ) e^{-t\left ( \frac{s}{c}\right ) }dt\) to converge. Which is what we required to show.
From definition\begin{align*} \mathcal{L}\left \{ \frac{1}{k}f\left ( \frac{t}{k}\right ) \right \} & =\frac{1}{k}\mathcal{L}\left \{ f\left ( \frac{t}{k}\right ) \right \} \\ & =\frac{1}{k}\int _{0}^{\infty }f\left ( \frac{t}{k}\right ) e^{-st}dt \end{align*}
Let \(\frac{t}{k}=\tau \). When \(t=0,\tau =0\) and when \(t=\infty ,\tau =\infty \). \(\frac{dt}{d\tau }=k\), hence the above becomes\begin{align*} \mathcal{L}\left \{ \frac{1}{k}f\left ( \frac{t}{k}\right ) \right \} & =\frac{1}{k}\int _{0}^{\infty }f\left ( \tau \right ) e^{-s\left ( k\tau \right ) }\left ( kd\tau \right ) \\ & =\int _{0}^{\infty }f\left ( \tau \right ) e^{-\tau \left ( sk\right ) }d\tau \end{align*}
We see from above that \(\mathcal{L}\left \{ \frac{1}{k}f\left ( \frac{t}{k}\right ) \right \} \) is \(F\left ( sk\right ) \). In other words, \(\mathcal{L}^{-1}\left \{ F\left ( ks\right ) \right \} =\frac{1}{k}f\left ( \frac{t}{k}\right ) \).
From definition\begin{align*} \mathcal{L}\left \{ \frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} & =\frac{1}{a}\mathcal{L}\left \{ e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} \\ & =\frac{1}{a}\int _{0}^{\infty }e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) e^{-st}dt \end{align*}
Let \(\frac{t}{a}=\tau \), at \(t=0,\tau =0\) and at \(t=\infty ,\tau =\infty \). And \(\frac{dt}{d\tau }=a\), hence the above becomes\begin{align*} \mathcal{L}\left \{ \frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} & =\frac{1}{a}\int _{0}^{\infty }e^{\frac{-b\left ( a\tau \right ) }{a}}f\left ( \tau \right ) e^{-s\left ( a\tau \right ) }\left ( ad\tau \right ) \\ & =\int _{0}^{\infty }e^{-b\tau }f\left ( \tau \right ) e^{-\tau \left ( sa\right ) }d\tau \\ & =\int _{0}^{\infty }f\left ( \tau \right ) e^{-\tau \left ( sa+b\right ) }d\tau \end{align*}
We see from the above, that \(\mathcal{L}\left \{ \frac{1}{a}e^{\frac{-bt}{a}}f\left ( \frac{t}{a}\right ) \right \} =F\left ( sa+b\right ) \). Now we look at the conditions which makes the above integral converges. Let \[ \left \vert f\left ( \tau \right ) e^{-t\left ( sa+b\right ) }\right \vert \leq k\left \vert e^{at}e^{-t\left ( sa+b\right ) }\right \vert \] Where \(k\) is some constant. Then\begin{align*} \int _{0}^{\infty }f\left ( t\right ) e^{-t\left ( sa+b\right ) }dt & \leq k\int _{0}^{\infty }e^{at}e^{-t\left ( sa+b\right ) }dt\\ & =k\int _{0}^{\infty }e^{-t\left ( sa+b-a\right ) }dt \end{align*}
But \(\int _{0}^{\infty }e^{-t\left ( sa+b-a\right ) }dt\) converges if \(sa+b-a>0\) or \(sa>a-b\) or \(s>1-\frac{b}{a}\)
Find inverse Laplace transform of \(F\left ( s\right ) =\frac{2^{n+1}n!}{s^{n+1}}\)
Solution
We know from tables that \[ \frac{n!}{s^{n+1}}\Longleftrightarrow t^{n}\] Hence\begin{align*} 2^{n+1}\frac{n!}{s^{n+1}} & \Longleftrightarrow 2^{n+1}t^{n}\\ & =2\left ( 2t\right ) ^{n} \end{align*}
Find inverse Laplace transform of \(F\left ( s\right ) =\frac{2s+1}{4s^{2}+4s+5}\)
Solution \[ F\left ( s\right ) =\frac{2s}{4s^{2}+4s+5}+\frac{1}{4s^{2}+4s+5}\] But \(4s^{2}+4s+5=4\left ( s+\frac{1}{2}\right ) ^{2}+4\), hence\begin{align} F\left ( s\right ) & =\frac{2s}{4\left ( s+\frac{1}{2}\right ) ^{2}+4}+\frac{1}{4\left ( s+\frac{1}{2}\right ) ^{2}+4}\nonumber \\ & =\frac{s}{2\left ( s+\frac{1}{2}\right ) ^{2}+2}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s}{\left ( s+\frac{1}{2}\right ) ^{2}+1}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s+\frac{1}{2}-\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s+\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1}-\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}+\frac{1}{4}\frac{1}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\nonumber \\ & =\frac{1}{2}\frac{s+\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1} \tag{1} \end{align}
Now we ready to do the inversion. Using \(e^{-at}f\left ( t\right ) \Longleftrightarrow F\left ( s+a\right ) \) and using \(\sin \left ( at\right ) \Longleftrightarrow \frac{a}{s^{2}+a^{2}}\), and \(\cos \left ( at\right ) \Longleftrightarrow \frac{s}{s^{2}+a^{2}}\)then\[ \frac{1}{2}\frac{s+\frac{1}{2}}{\left ( s+\frac{1}{2}\right ) ^{2}+1}\Longleftrightarrow \frac{1}{2}e^{-\frac{1}{2}t}\cos \left ( t\right ) \] Hence\[ f\left ( t\right ) =\frac{1}{2}e^{-\frac{1}{2}t}\cos \left ( t\right ) \]
Find inverse Laplace transform of \(F\left ( s\right ) =\frac{1}{9s^{2}-12s+3}\)
Solution\[ \frac{1}{9s^{2}-12s+3}=\frac{1}{9}\frac{1}{s^{2}-\frac{4}{3}s+\frac{1}{3}}=\frac{1}{9}\frac{1}{\left ( s-1\right ) \left ( s-\frac{1}{3}\right ) }\] But \begin{align*} \frac{1}{\left ( s-1\right ) \left ( s-\frac{1}{3}\right ) } & =\frac{A}{s-1}+\frac{B}{s-\frac{1}{3}}\\ A & =\left ( \frac{1}{\left ( s-\frac{1}{3}\right ) }\right ) _{s=1}=\frac{3}{2}\\ B & =\left ( \frac{1}{\left ( s-1\right ) }\right ) _{s=\frac{1}{3}}=-\frac{3}{2} \end{align*}
Hence\begin{equation} \frac{1}{9s^{2}-12s+3}=\frac{1}{9}\left ( \frac{3}{2}\frac{1}{s-1}-\frac{3}{2}\frac{1}{s-\frac{1}{3}}\right ) \tag{1} \end{equation} Using \[ e^{at}\Longleftrightarrow \frac{1}{s-a}\] Then (1) becomes\begin{align*} \frac{1}{9s^{2}-12s+3} & \Longleftrightarrow \frac{1}{9}\left ( \frac{3}{2}e^{t}-\frac{3}{2}e^{\frac{1}{3}t}\right ) \\ & =\frac{1}{6}e^{t}-\frac{1}{6}e^{\frac{1}{3}t}\\ & =\frac{1}{6}\left ( e^{t}-e^{\frac{1}{3}t}\right ) \end{align*}
Find inverse Laplace transform of \(F\left ( s\right ) =\frac{e^{2}e^{-4s}}{2s-1}\)
solution
\[ F\left ( s\right ) =\frac{e^{2}}{2}\frac{e^{-4s}}{s-\frac{1}{2}}\] Using \begin{equation} u_{c}\left ( t\right ) f\left ( t-c\right ) \Longleftrightarrow e^{-cs}F\left ( s\right ) \tag{1} \end{equation} Since \[ \frac{1}{s-\frac{1}{2}}\Longleftrightarrow e^{\frac{1}{2}t}\] Then using (1)\[ e^{-4s}\frac{1}{s-\frac{1}{2}}\Longleftrightarrow u_{4}\left ( t\right ) e^{\frac{1}{2}\left ( t-4\right ) }\] Hence\begin{align*} \frac{e^{2}}{2}\frac{e^{-4s}}{s-\frac{1}{2}} & \Longleftrightarrow \frac{e^{2}}{2}u_{4}\left ( t\right ) e^{\frac{1}{2}\left ( t-4\right ) }\\ & =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{1}{2}\left ( t-4\right ) +2}\\ & =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{1}{2}t-2+2}\\ & =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{t}{2}} \end{align*}
Therefore\[ f\left ( t\right ) =\frac{1}{2}u_{4}\left ( t\right ) e^{\frac{t}{2}}\] ps. Book answer is wrong. It gives\[ f\left ( t\right ) =\frac{1}{2}u_{4}\left ( \frac{t}{2}\right ) e^{\frac{t}{2}}\]
Find Laplace transform of \(f\left ( t\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq t<1\\ 0 & & t\geq 1 \end{array} \right . \)
solution
Writing \(f\left ( t\right ) \) in terms of Heaviside step function gives\[ f\left ( t\right ) =u_{0}\left ( t\right ) -u_{1}\left ( t\right ) \] Using\[ u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}\] Therefore\begin{align*} \mathcal{L}\left \{ u_{0}\left ( t\right ) \right \} & =e^{-0s}\frac{1}{s}=\frac{1}{s}\\\mathcal{L}\left \{ u_{1}\left ( t\right ) \right \} & =e^{-s}\frac{1}{s} \end{align*}
Hence\begin{align*} \mathcal{L}\left \{ u_{0}\left ( t\right ) -u_{1}\left ( t\right ) \right \} & =\frac{1}{s}-e^{-s}\frac{1}{s}\\ & =\frac{1}{s}\left ( 1-e^{-s}\right ) \qquad s>0 \end{align*}
Find Laplace transform of \(f\left ( t\right ) =\left \{ \begin{array} [c]{ccc}1 & & 0\leq t<1\\ 0 & & 1\leq t<2\\ 1 & & 2\leq t<3\\ 0 & & t\geq 3 \end{array} \right . \)
solution
Writing \(f\left ( t\right ) \) in terms of Heaviside step function gives\[ f\left ( t\right ) =u_{0}\left ( t\right ) -u_{1}\left ( t\right ) +u_{2}\left ( t\right ) -u_{3}\left ( t\right ) \] Using\[ u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}\] But \(f\left ( t\right ) =1\) in this case. Hence \(F\left ( s\right ) =\frac{1}{s}\). Therefore\begin{align*} f\left ( t\right ) & \Longleftrightarrow \frac{1}{s}e^{0s}-\frac{1}{s}e^{-s}+\frac{1}{s}e^{-2s}-\frac{1}{s}e^{-3s}\\ & =\frac{1}{s}\left ( 1-e^{-s}+e^{-2s}-e^{-3s}\right ) \qquad s>0 \end{align*}
Find Laplace transform of \(f\left ( t\right ) =1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \)
solution
Using\[ u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}\] Therefore\begin{align*} \mathcal{L}\left \{ 1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} & =\mathcal{L}\left \{ 1\right \} +\mathcal{L}\left \{ \sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} \\ & =\frac{1}{s}+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}\frac{1}{s}e^{-ks}\\ & =\sum _{k=0}^{2n+1}\left ( -1\right ) ^{k}\frac{1}{s}e^{-ks}\\ & =\frac{1}{s}\sum _{k=0}^{2n+1}\left ( -e^{-s}\right ) ^{k} \end{align*}
Since \(\left \vert e^{-s}\right \vert <1\) the sum converges. Using \(\sum _{0}^{N}a_{n}=\left ( \frac{1-r^{N+1}}{1-r}\right ) \). Where \(\left \vert r\right \vert <1\). So the answer is\begin{align*} \mathcal{L}\left \{ 1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} & =\frac{1}{s}\left ( \frac{1-\left ( -e^{-s}\right ) ^{2n+2}}{1-\left ( -e^{-s}\right ) }\right ) \\ & =\frac{1}{s}\left ( \frac{1-\left ( -e\right ) ^{-\left ( 2n+2\right ) s}}{1+e^{-s}}\right ) \end{align*}
Since \(2n+2\) is even then\[\mathcal{L}\left \{ 1+\sum _{k=1}^{2n+1}\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} =\frac{1}{s}\left ( \frac{1+e^{-\left ( 2n+2\right ) s}}{1+e^{-s}}\right ) \qquad s>0 \]
Find Laplace transform of \(f\left ( t\right ) =1+\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \)
solution
Using\[ u_{c}\left ( t\right ) \Longleftrightarrow e^{-cs}\frac{1}{s}\] Therefore\begin{align*} \mathcal{L}\left \{ 1+\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} & =\mathcal{L}\left \{ 1\right \} +\mathcal{L}\left \{ \sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k}\left ( t\right ) \right \} \\ & =\frac{1}{s}+\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}\frac{1}{s}e^{-ks}\\ & =\frac{1}{s}+\frac{1}{s}\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}e^{-ks}\\ & =\frac{1}{s}+\frac{1}{s}\sum _{k=1}^{\infty }\left ( -e^{-s}\right ) ^{k} \end{align*}
But \[ \sum _{k=1}^{\infty }r^{k}=\frac{r}{1-r}\qquad \left \vert r\right \vert <1 \] Since \(s>0\) then\(\left \vert e^{-s}\right \vert <1\). So the answer is\begin{align*} \frac{1}{s}+\frac{1}{s}\frac{-e^{-s}}{1-\left ( -e^{-s}\right ) } & =\frac{1}{s}-\frac{1}{s}\frac{e^{-s}}{1+e^{-s}}\\ & =\frac{1+e^{-s}-e^{-s}}{s\left ( 1+e^{-s}\right ) }\\ & =\frac{1}{s\left ( 1+e^{-s}\right ) }\qquad s>0 \end{align*}
A plot of part (a) is the following
And a plot of part(a) for problem 19 is the following
We see the effect of having a \(2\) inside the sum. It extends the step \(u_{c}\left ( t\right ) \) function to negative side.
The easy way to do this, is to solve for each input term separately, and then add all the solutions, since this is a linear ODE. Once we solve for the first 2-3 terms, we will see the pattern to use for the overall solution. Since the input \(g\left ( t\right ) \) is \(u_{0}\left ( t\right ) +\sum _{k=1}^{\infty }\left ( -1\right ) ^{k}u_{k\pi }\left ( t\right ) \), we will first first the response to \(u_{0}\left ( t\right ) \,\), then for \(-u_{\pi }\left ( t\right ) \) then for \(+u_{2\pi }\left ( t\right ) \), and so on, and add them.
When the input is \(u_{0}\left ( t\right ) \), then its Laplace transform is \(\frac{1}{s}\), Hence, taking Laplace transform of the ODE gives (where now \(Y\left ( s\right ) =\mathcal{L}\left ( y\left ( t\right ) \right ) \))\[ \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) +y^{\prime }\left ( 0\right ) \right ) +Y\left ( s\right ) =\frac{1}{s}\] Applying initial conditions\[ s^{2}Y\left ( s\right ) +Y\left ( s\right ) =\frac{1}{s}\] Solving for \(Y_{0}\left ( s\right ) \) (called it \(Y_{0}\left ( s\right ) \,\) since the input is \(u_{0}\left ( t\right ) \))\begin{align*} Y_{0}\left ( s\right ) & =\frac{1}{s\left ( s^{2}+1\right ) }\\ & =\frac{1}{s}-\frac{s}{s^{2}+1} \end{align*}
Hence\[ y_{0}\left ( t\right ) =1-\cos t \] We now do the next input, which is \(-u_{\pi }\left ( t\right ) \), which has Laplace transform of \(-\frac{e^{-\pi s}}{s}\), therefore, following what we did above, we obtain now\begin{align*} Y_{\pi }\left ( s\right ) & =\frac{-e^{-\pi s}}{s\left ( s^{2}+1\right ) }\\ & =-e^{-\pi s}\left ( \frac{1}{s}-\frac{s}{s^{2}+1}\right ) \end{align*}
The effect of \(e^{-\pi s}\) is to cause delay in time. Hence the the inverse Laplace transform of the above is the same as \(y_{0}\left ( t\right ) \) but with delay \[ y_{\pi }\left ( t\right ) =-u_{\pi }\left ( t\right ) \left ( 1-\cos \left ( t-\pi \right ) \right ) \] Similarly, when the input is \(+u_{2\pi }\left ( t\right ) \), which which has Laplace transform of \(\frac{e^{-2\pi s}}{s}\), therefore, following what we did above, we obtain now\begin{align*} Y_{\pi }\left ( s\right ) & =\frac{e^{-2\pi s}}{s\left ( s^{2}+1\right ) }\\ & =e^{-2\pi s}\left ( \frac{1}{s}-\frac{s}{s^{2}+1}\right ) \end{align*}
The effect of \(e^{-2\pi s}\) is to cause delay in time. Hence the the inverse Laplace transform of the above is the same as \(y_{0}\left ( t\right ) \) but with now with delay of \(2\pi \), therefore\[ y_{2\pi }\left ( t\right ) =+u_{2\pi }\left ( t\right ) \left ( 1-\cos \left ( t-2\pi \right ) \right ) \] And so on. We see that if we add all the responses, we obtain\begin{align*} y\left ( t\right ) & =y_{0}\left ( t\right ) +y_{\pi }\left ( t\right ) +y_{2\pi }\left ( t\right ) +\cdots \\ & =\left ( 1-\cos t\right ) -u_{\pi }\left ( t\right ) \left ( 1-\cos \left ( t-\pi \right ) \right ) +u_{2\pi }\left ( t\right ) \left ( 1-\cos \left ( t-2\pi \right ) \right ) -\cdots \end{align*}
Or\begin{equation} y\left ( t\right ) =\left ( 1-\cos t\right ) +\sum _{k=1}^{n}\left ( -1\right ) ^{k}u_{k\pi }\left ( t\right ) \left ( 1-\cos \left ( t-k\pi \right ) \right ) \tag{1} \end{equation}
This is a plot of (1) for \(n=15\)
We see the solution growing rapidly, they settling down after about \(t=50\) to sinusoidal wave at amplitude of about \(\pm 15\). This shows the system reached steady state at around \(t=50\).
To compare it with problem 19 solution, I used the solution for 19 given in the book, and plotted both solution on top of each others. Also for up to \(t=60\). Here is the result
We see that problem 19 output follows the same pattern (since same frequency is used), but with double the amplitude. This is due to the \(2\) factor used in problem 19 compared to this problem.
At first, I tried it with \(n=50,150,250,350,450,550.\) I can not see any noticeable change in the plot. Here is the result.
Even at \(n=2000\) there was no change to be noticed.
This shows additional input in the form of shifted unit steps, do not change the steady state solution.