Use the Laplace transform to solve \(y^{\prime \prime }-4y^{\prime }+4y=t^{3}e^{2t}\) with IC \(y\left ( 0\right ) =0\) and \(y^{\prime }\left ( 0\right ) =0\)
Solution:
Let \(Y\left ( s\right ) =\left ( y\left ( t\right ) \right ) \). Taking Laplace transform of the above ODE gives\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -4\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +4Y\left ( s\right ) =\mathcal{L}\left ( t^{3}e^{2t}\right ) \tag{1} \end{equation} But \(\mathcal{L}\left ( t^{n}e^{at}\right ) =\left ( -1\right ) ^{n}\frac{d^{n}}{ds^{n}}\left ( \mathcal{L}\left ( e^{at}\right ) \right ) \), where \(\mathcal{L}\left ( e^{at}\right ) =\frac{1}{s-a}\), therefore\begin{align*} \mathcal{L}\left ( t^{3}e^{2t}\right ) & =\left ( -1\right ) ^{3}\frac{d^{3}}{ds^{3}}\left ( \mathcal{L}\left ( e^{2t}\right ) \right ) \\ & =\left ( -1\right ) ^{3}\frac{d^{3}}{ds^{3}}\left ( \frac{1}{s-2}\right ) \\ & =\frac{6}{\left ( s-2\right ) ^{4}} \end{align*}
Eq. (1) becomes\begin{align*} s^{2}Y\left ( s\right ) -4sY\left ( s\right ) +4Y\left ( s\right ) & =\frac{6}{\left ( s-2\right ) ^{4}}\\ Y\left ( s\right ) \left ( s^{2}-4s+4\right ) & =\frac{6}{\left ( s-2\right ) ^{4}}\\ Y\left ( s\right ) & =\frac{6}{\left ( s-2\right ) ^{4}\left ( s^{2}-4s+4\right ) }\\ & =\frac{6}{\left ( s-2\right ) ^{4}\left ( s-2\right ) ^{2}}\\ & =\frac{6}{\left ( s-2\right ) ^{6}} \end{align*}
Using the property \(\mathcal{L}^{-1}\left ( \frac{1}{\left ( s-a\right ) ^{n}}\right ) =e^{at}\mathcal{L}^{-1}\left ( \frac{1}{s^{n}}\right ) \,\), the above reduces to\[ y\left ( t\right ) =6e^{2t}\mathcal{L}^{-1}\left ( \frac{1}{s^{6}}\right ) \] Since \(\mathcal{L}^{-1}\left ( \frac{1}{s^{n+1}}\right ) =\frac{t^{n}}{n!}\), hence \(\left ( \frac{1}{s^{6}}\right ) =\frac{t^{5}}{120}\), and the above becomes\[ y\left ( t\right ) =\frac{1}{20}t^{5}e^{2t}\]
Use Laplace transform to solve
\(y^{\prime \prime }-4y^{\prime }+3y=1-H\left ( t-2\right ) -H\left ( t-4\right ) +H\left ( t-6\right ) \) with IC \(y\left ( 0\right ) =0\) and \(y^{\prime }\left ( 0\right ) =0\)
solution:
Let \(Y\left ( s\right ) =\mathcal{L}\left ( y\left ( t\right ) \right ) \). Taking Laplace transform of the above ODE gives\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -4\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +3Y\left ( s\right ) =\mathcal{L}\left ( 1-H\left ( t-2\right ) -H\left ( t-4\right ) +H\left ( t-6\right ) \right ) \tag{1} \end{equation} But \(\mathcal{L}\left ( 1\right ) =\frac{1}{s}\) and \(\mathcal{L}\left ( H\left ( t-a\right ) \right ) =\frac{e^{-as}}{s}\), hence the above becomes\begin{align*} s^{2}Y\left ( s\right ) -4sY\left ( s\right ) +3Y\left ( s\right ) & =\frac{1}{s}-\frac{e^{-2s}}{s}-\frac{e^{-4s}}{s}+\frac{e^{-6s}}{s}\\ Y\left ( s\right ) & =\frac{1-e^{-2s}-e^{-4s}+e^{-6s}}{s\left ( s^{2}-4s+3\right ) }\\ & =\frac{1-e^{-2s}-e^{-4s}+e^{-6s}}{s\left ( s-3\right ) \left ( s-1\right ) } \end{align*}
The inverse Laplace transform of \(\frac{1}{s\left ( s-3\right ) \left ( s-1\right ) }\) is found first. Let the result be \(f\left ( t\right ) \). Then the relation \(\mathcal{L}^{-1}\left \{ e^{-as}F\left ( s\right ) \right \} =f\left ( t-a\right ) H\left ( t-a\right ) \) is used to obtain the final answer. Using partial fractions. Let \[ \frac{1}{s\left ( s-3\right ) \left ( s-1\right ) }=\frac{A}{s}+\frac{B}{s-3}+\frac{C}{s-1}\] Then \(A=\lim _{s\rightarrow 0}\frac{1}{\left ( s-3\right ) \left ( s-1\right ) }=\frac{1}{3}\) and \(B=\lim _{s\rightarrow 3}\frac{1}{s\left ( s-1\right ) }=\frac{1}{6}\) and \(C=\lim _{s\rightarrow 1}\frac{1}{s\left ( s-3\right ) }=-\frac{1}{2}\), hence\begin{align*} \mathcal{L}^{-1}\left \{ \frac{1}{s\left ( s-3\right ) \left ( s-1\right ) }\right \} & =\mathcal{L}^{-1}\left \{ \frac{1}{3}\frac{1}{s}+\frac{1}{6}\frac{1}{s-3}-\frac{1}{2}\frac{1}{s-1}\right \} \\ & =\frac{1}{3}+\frac{1}{6}e^{3t}-\frac{1}{2}e^{t} \end{align*}
The above is \(f\left ( t\right ) \). Therefore, using \(\mathcal{L}^{-1}\left \{ e^{-as}F\left ( s\right ) \right \} =f\left ( t-a\right ) H\left ( t-a\right ) \)\begin{align*} y\left ( t\right ) & =\mathcal{L}^{-1}\left ( 1-e^{-2s}-e^{-4s}+e^{-6s}\right ) F\left ( s\right ) \\ & =f\left ( t\right ) -f\left ( t-2\right ) H\left ( t-2\right ) -f\left ( t-4\right ) H\left ( t-4\right ) +f\left ( t-6\right ) H\left ( t-6\right ) \end{align*}
Hence the answer is\begin{align*} y\left ( t\right ) & =\left ( \frac{1}{3}+\frac{1}{6}e^{3t}-\frac{1}{2}e^{t}\right ) -\left ( \frac{1}{3}+\frac{1}{6}e^{3\left ( t-2\right ) }-\frac{1}{2}e^{\left ( t-2\right ) }\right ) H\left ( t-2\right ) \\ & -\left ( \frac{1}{3}+\frac{1}{6}e^{3\left ( t-4\right ) }-\frac{1}{2}e^{\left ( t-4\right ) }\right ) H\left ( t-4\right ) +\left ( \frac{1}{3}+\frac{1}{6}e^{3\left ( t-6\right ) }-\frac{1}{2}e^{\left ( t-6\right ) }\right ) H\left ( t-6\right ) \end{align*}
Use Laplace transform to solve \(y^{\prime \prime }-4y^{\prime }+13y=\delta \left ( t-\pi \right ) +\delta \left ( t-3\pi \right ) \) with IC \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0\)
Solution:
The property \(\mathcal{L}\left ( \delta \left ( t-a\right ) \right ) =e^{-as}\) will be used. Taking the Laplace transform of the ODE gives\begin{equation} \left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -4\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +13Y\left ( s\right ) =e^{-\pi s}+e^{-3\pi s} \tag{1} \end{equation} Hence\begin{align*} \left ( s^{2}Y\left ( s\right ) -s\right ) -4\left ( sY\left ( s\right ) -1\right ) +13Y\left ( s\right ) & =e^{-\pi s}+e^{-3\pi s}\\ \left [ s^{2}Y\left ( s\right ) -4sY\left ( s\right ) +13Y\left ( s\right ) \right ] +\left ( -s+4\right ) & =e^{-\pi s}+e^{-3\pi s}\\ Y\left ( s\right ) \left ( s^{2}-4s+13\right ) & =e^{-\pi s}+e^{-3\pi s}+\left ( s-4\right ) \\ Y\left ( s\right ) & =\frac{e^{-\pi s}+e^{-3\pi s}+\left ( s-4\right ) }{s^{2}-4s+13}\\ & =\frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13}+\frac{\left ( s-4\right ) }{s^{2}-4s+13}\\ & =\frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13}+\frac{\left ( s-2\right ) -2}{\left ( s-2\right ) ^{2}+9} \end{align*}
To find the inverse Laplace of \(\frac{\left ( s-2\right ) -2}{\left ( s-2\right ) ^{2}+9}\) the following property is used \(\mathcal{L}^{-1}\left ( \frac{\left ( s-a\right ) }{\left ( s-a\right ) ^{n}+b}\right ) =e^{at}\mathcal{L}^{-1}\left ( \frac{s}{s^{n}+b}\right ) \), Hence\begin{align} \mathcal{L}^{-1}\left ( \frac{\left ( s-2\right ) -2}{\left ( s-2\right ) ^{2}+9}\right ) & =e^{2t}\mathcal{L}^{-1}\left ( \frac{s-2}{s^{2}+9}\right ) \nonumber \\ & =e^{2t}\left [ \mathcal{L}^{-1}\left ( \frac{s}{s^{2}+9}\right ) -2\mathcal{L}^{-1}\left ( \frac{1}{s^{2}+9}\right ) \right ] \nonumber \\ & =e^{2t}\left [ \mathcal{L}^{-1}\left ( \frac{s}{s^{2}+3^{2}}\right ) -\frac{2}{3}\mathcal{L}^{-1}\left ( \frac{3}{s^{2}+3^{2}}\right ) \right ] \nonumber \\ & =e^{2t}\left ( \cos 3t-\frac{2}{3}\sin 3t\right ) \nonumber \\ & =\frac{1}{3}e^{2t}\left ( 3\cos 3t-2\sin 3t\right ) \tag{2} \end{align}
Now the inverse Laplace transform of \(\frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13}\) is found. Writing this as\begin{align*} \frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13} & =\frac{e^{-\pi s}}{s^{2}-4s+13}+\frac{e^{-3\pi s}}{s^{2}-4s+13}\\ & =\frac{e^{-\pi s}}{\left ( s-2\right ) ^{2}+9}+\frac{e^{-3\pi s}}{\left ( s-2\right ) ^{2}+9} \end{align*}
To be able to use the property \(\mathcal{L}^{-1}\left ( \frac{F\left ( s-a\right ) }{\left ( s-a\right ) ^{n}+b}\right ) =e^{at}\mathcal{L}^{-1}\left ( \frac{F\left ( s\right ) }{s^{n}+b}\right ) \) the terms in the numerator are converted as follows\[ \frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13}=e^{-2\pi }\frac{e^{-\pi \left ( s-2\right ) }}{\left ( s-2\right ) ^{2}+9}+e^{-6\pi }\frac{e^{-3\pi \left ( s-2\right ) }}{\left ( s-2\right ) ^{2}+9}\] Now the property can be used, hence\begin{align*} \mathcal{L}^{-1}\left ( \frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13}\right ) & =e^{-2\pi }e^{2t}\mathcal{L}^{-1}\left ( \frac{e^{-\pi s}}{s^{2}+9}\right ) +e^{-6\pi }e^{2t}\mathcal{L}^{-1}\left ( \frac{e^{-3\pi s}}{s^{2}+9}\right ) \\ & =e^{2t}\left ( e^{-2\pi }\mathcal{L}^{-1}\left ( \frac{e^{-\pi s}}{s^{2}+9}\right ) +e^{-6\pi }\mathcal{L}^{-1}\left ( \frac{e^{-3\pi s}}{s^{2}+9}\right ) \right ) \end{align*}
Now another property is used to find \(\mathcal{L}^{-1}\) of the remaining terms. This is \(\mathcal{L}^{-1}\left ( e^{-as}F\left ( s\right ) \right ) =f\left ( t-a\right ) H\left ( t-a\right ) \). The above becomes\begin{align} \mathcal{L}^{-1}\left ( \frac{e^{-\pi s}+e^{-3\pi s}}{s^{2}-4s+13}\right ) & =e^{2t}\left ( e^{-2\pi }\mathcal{L}^{-1}\left ( \frac{1}{s^{2}+3^{2}}\right ) H\left ( t-\pi \right ) +e^{-6\pi }\mathcal{L}^{-1}\left ( \frac{1}{s^{2}+3^{2}}\right ) H\left ( t-3\pi \right ) \right ) \nonumber \\ & =e^{2t}\left ( \frac{1}{3}e^{-2\pi }\mathcal{L}^{-1}\left ( \frac{3}{s^{2}+3^{2}}\right ) H\left ( t-\pi \right ) +\frac{1}{3}e^{-6\pi }\mathcal{L}^{-1}\left ( \frac{3}{s^{2}+3^{2}}\right ) H\left ( t-3\pi \right ) \right ) \nonumber \\ & =e^{2t}\left ( \frac{1}{3}e^{-2\pi }\sin \left ( 3\left ( t-\pi \right ) \right ) H\left ( t-\pi \right ) +\frac{1}{3}e^{-6\pi }\sin \left ( 3\left ( t-3\pi \right ) \right ) H\left ( t-3\pi \right ) \right ) \tag{3} \end{align}
The full solution is now found, combining Eq. (2) and Eq. (3) gives\[ y\left ( t\right ) =-\frac{1}{3}e^{2t}\left ( 3\cos 3t-2\sin 3t\right ) +e^{2t}\left ( \frac{1}{3}e^{-2\pi }\sin \left ( 3\left ( t-\pi \right ) \right ) H\left ( t-\pi \right ) +\frac{1}{3}e^{-6\pi }\sin \left ( 3\left ( t-3\pi \right ) \right ) H\left ( t-3\pi \right ) \right ) \] Taking common factors out gives\[ y\left ( t\right ) =-\frac{1}{3}e^{2t}\left ( 2\sin \left ( 3t\right ) -3\cos \left ( 3t\right ) +e^{-2\pi }\sin \left ( 3\left ( t-\pi \right ) \right ) H\left ( t-\pi \right ) +e^{-6\pi }\sin \left ( 3\left ( t-3\pi \right ) \right ) H\left ( t-3\pi \right ) \right ) \] The above can be reduced more. Since \(\sin \left ( n\left ( t-\pi \right ) \right ) =-1^{n}\sin nt\) for integer \(n\), hence the above simplifies to\[ y\left ( t\right ) =-\frac{1}{3}e^{2t}\left ( 2\sin \left ( 3t\right ) -3\cos \left ( 3t\right ) -e^{-2\pi }\sin \left ( 3t\right ) H\left ( t-\pi \right ) -e^{-6\pi }\sin \left ( 3t\right ) H\left ( t-3\pi \right ) \right ) \] For \(t\geq 0\).
Solve the boundary valued problem using Laplace transform
\(ty^{\prime \prime }-\left ( t+3\right ) y^{\prime }+4y=t-1\) where \(y\left ( 0\right ) =y\left ( 1\right ) =0\)
Solution:
The following property will be used \(\mathcal{L}\left ( t^{n}f\left ( t\right ) \right ) =\left ( -1\right ) ^{n}\frac{d^{n}}{ds^{n}}\left ( \mathcal{L}\left ( f\left ( t\right ) \right ) \right ) \). Taking Laplace transform of the above gives\begin{align*} -\frac{d}{ds}\left ( s^{2}Y\left ( s\right ) -sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +\frac{d}{ds}\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) -3\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +4Y\left ( s\right ) & =\frac{1}{s^{2}}-\frac{1}{s}\\ -\frac{d}{ds}\left ( s^{2}Y\left ( s\right ) -y^{\prime }\left ( 0\right ) \right ) +\frac{d}{ds}sY\left ( s\right ) -3sY\left ( s\right ) +4Y\left ( s\right ) & =\frac{1}{s^{2}}-\frac{1}{s}\\ -\left ( 2sY\left ( s\right ) +s^{2}Y^{\prime }\left ( s\right ) \right ) +\left ( Y\left ( s\right ) +sY^{\prime }\left ( s\right ) \right ) -3sY\left ( s\right ) +4Y\left ( s\right ) & =\frac{1}{s^{2}}-\frac{1}{s}\\ -2sY\left ( s\right ) -s^{2}Y^{\prime }\left ( s\right ) +Y\left ( s\right ) +sY^{\prime }\left ( s\right ) -3sY\left ( s\right ) +4Y\left ( s\right ) & =\frac{1}{s^{2}}-\frac{1}{s}\\ Y^{\prime }\left ( s\right ) \left ( -s^{2}+s\right ) -5sY\left ( s\right ) +5Y\left ( s\right ) & =\frac{1}{s^{2}}-\frac{1}{s}\\ Y^{\prime }\left ( s\right ) -\frac{5s}{\left ( -s^{2}+s\right ) }Y\left ( s\right ) +\frac{5}{\left ( -s^{2}+s\right ) }Y\left ( s\right ) & =\frac{\frac{1}{s^{2}}-\frac{1}{s}}{\left ( -s^{2}+s\right ) }\\ Y^{\prime }\left ( s\right ) +\frac{5s}{\left ( s^{2}-s\right ) }Y\left ( s\right ) -\frac{5}{\left ( s^{2}-s\right ) }Y\left ( s\right ) & =\frac{1-s}{s^{2}\left ( -s^{2}+s\right ) }\\ Y^{\prime }\left ( s\right ) +\frac{5}{s}Y\left ( s\right ) & =\frac{1}{s^{3}} \end{align*}
Integrating factor is \(\ln \left ( I_{f}\right ) =\int \frac{5}{s}ds\), hence \(I_{f}=e^{5\int \frac{1}{s}ds}=e^{5\ln s}=s^{5}\), therefore\begin{align*} d\left ( I_{f}Y\right ) & =s^{5}\frac{1}{s^{3}}\\ I_{f}Y & =\int s^{2}ds+c\\ & =\frac{s^{3}}{3}+c \end{align*}
Then\[ Y\left ( s\right ) =\frac{1}{3s^{2}}+\frac{c}{s^{5}}\] Using the property \(\mathcal{L}^{-1}\left ( \frac{n!}{s^{n+1}}\right ) =t^{n}\) for \(n=1,2,3,\cdots \) then \(\mathcal{L}^{-1}\left ( \frac{1}{s^{n+1}}\right ) =\frac{t^{n}}{n!}\), therefore taking the inverse Laplace transform of the above gives\begin{align*} y\left ( t\right ) & =\frac{1}{3}t+c\frac{t^{4}}{4!}\\ & =\frac{1}{3}t+c\frac{t^{4}}{24} \end{align*}
To find the constant \(c\,\), from the second boundary condition\begin{align*} y\left ( 1\right ) & =0\\ \frac{1}{3}+c\frac{1}{24} & =0\\ c & =-\frac{24}{3}=-8 \end{align*}
Hence\[ y\left ( t\right ) =\frac{1}{3}t\left ( 1-t^{3}\right ) \]
Solve the following system of equations for \(y\left ( t\right ) \) and \(z\left ( t\right ) \)\begin{align*} 3y^{\prime }+8y+2z^{\prime }+5z & =e^{-t}\\ y^{\prime }+z^{\prime }+z & =0 \end{align*}
Where the IC are \(y\left ( 0\right ) =2\) and \(z\left ( 0\right ) =-2\)
Solution:
Taking Laplace transform of the system of equation gives\begin{align*} 3\left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +8Y\left ( s\right ) +2\left ( sZ\left ( s\right ) -z\left ( 0\right ) \right ) +5Z\left ( s\right ) & =\frac{1}{s+1}\\ \left ( sY\left ( s\right ) -y\left ( 0\right ) \right ) +\left ( sZ\left ( s\right ) -z\left ( 0\right ) \right ) +Z\left ( s\right ) & =0 \end{align*}
Applying IC gives\begin{align*} 3\left ( sY\left ( s\right ) -2\right ) +8Y\left ( s\right ) +2\left ( sZ\left ( s\right ) +2\right ) +5Z\left ( s\right ) & =\frac{1}{s+1}\\ \left ( sY\left ( s\right ) -2\right ) +\left ( sZ\left ( s\right ) +2\right ) +Z\left ( s\right ) & =0 \end{align*}
Simplifying in order to solve for \(Y\left ( s\right ) \) and \(Z\left ( s\right ) \)\begin{align} Y\left ( s\right ) \left ( 3s+8\right ) +Z\left ( s\right ) \left ( 2s+5\right ) & =\frac{1}{s+1}+2\tag{1}\\ sY\left ( s\right ) +Z\left ( s\right ) \left ( 1+s\right ) & =0\nonumber \end{align}
From the second Eq. \[ Y\left ( s\right ) =-\frac{Z\left ( s\right ) \left ( 1+s\right ) }{s}\] Substituting this in the first equation above gives\[ -\frac{Z\left ( s\right ) \left ( 1+s\right ) }{s}\left ( 3s+8\right ) +Z\left ( s\right ) \left ( 2s+5\right ) =\frac{1}{s+1}+2 \] Simplifying\begin{align*} -Z\left ( s\right ) \left ( 1+s\right ) \left ( 3s+8\right ) +Z\left ( s\right ) s\left ( 2s+5\right ) & =\frac{s}{s+1}+2s\\ Z\left ( s\right ) \left [ -\left ( 1+s\right ) \left ( 3s+8\right ) +s\left ( 2s+5\right ) \right ] & =\frac{s+2s\left ( s+1\right ) }{s+1}\\ Z\left ( s\right ) \left [ -s^{2}-6s-8\right ] & =\frac{s+2s\left ( s+1\right ) }{s+1}\\ Z\left ( s\right ) & =-\frac{s\left ( 2s+3\right ) }{\left ( s+1\right ) \left ( s^{2}+6s+8\right ) }\\ & =-\frac{s\left ( 2s+3\right ) }{\left ( s+1\right ) \left ( s+4\right ) \left ( s+2\right ) } \end{align*}
Using Partial fractions, let \(\frac{s\left ( 2s+3\right ) }{\left ( s+1\right ) \left ( \allowbreak s+4\right ) \left ( s+2\right ) }=\frac{A}{\left ( s+1\right ) }+\frac{B}{\left ( s+4\right ) }+\frac{C}{s+2}\), hence\begin{align*} A & =\lim _{s\rightarrow -1}\frac{s\left ( 2s+3\right ) }{\left ( s+4\right ) \left ( s+2\right ) }=\frac{-1\left ( -2+3\right ) }{\left ( -1+4\right ) \left ( -1+2\right ) }=-\frac{1}{3}\\ B & =\lim _{s\rightarrow -4}\frac{s\left ( 2s+3\right ) }{\left ( s+1\right ) \left ( s+2\right ) }=\frac{-4\left ( -8+3\right ) }{\left ( -4+1\right ) \left ( -4+2\right ) }=\frac{10}{3}\\ C & =\lim _{s\rightarrow -2}\frac{s\left ( 2s+3\right ) }{\left ( s+1\right ) \left ( s+4\right ) }=\frac{-2\left ( -4+3\right ) }{\left ( -2+1\right ) \left ( -2+4\right ) }=-1 \end{align*}
Or\begin{align} Z\left ( s\right ) & =-\left ( \frac{A}{\left ( s+1\right ) }+\frac{B}{\left ( s+4\right ) }+\frac{C}{s+2}\right ) \nonumber \\ & =\frac{1}{3}\frac{1}{\left ( s+1\right ) }-\frac{10}{3}\frac{1}{\left ( s+4\right ) }+\frac{1}{s+2} \tag{2} \end{align}
Hence \[ z\left ( t\right ) =\frac{1}{3}e^{-t}-\frac{10}{3}e^{-4t}+e^{-2t}\] Now \(Y\left ( s\right ) \) will be found and solved for. Using the value for \(Z\left ( s\right ) \) from Eq. (2) and substituting this in Eq. (1) gives\begin{align*} Y\left ( s\right ) \left ( 3s+8\right ) +\left ( \frac{1}{3}\frac{1}{\left ( s+1\right ) }-\frac{10}{3}\frac{1}{\left ( s+4\right ) }+\frac{1}{s+2}\right ) \left ( 2s+5\right ) & =\frac{1}{s+1}+2\\ Y\left ( s\right ) \left ( 3s+8\right ) & =\frac{1}{s+1}+2-\frac{1}{3}\frac{\left ( 2s+5\right ) }{\left ( s+1\right ) }+\frac{10}{3}\frac{\left ( 2s+5\right ) }{\left ( s+4\right ) }-\frac{\left ( 2s+5\right ) }{s+2} \end{align*}
Hence\begin{align*} Y\left ( s\right ) & =\frac{1}{\left ( s+1\right ) \left ( 3s+8\right ) }+\frac{2}{\left ( 3s+8\right ) }-\frac{1}{3}\frac{\left ( 2s+5\right ) }{\left ( 3s+8\right ) \left ( s+1\right ) }+\frac{10}{3}\frac{\left ( 2s+5\right ) }{\left ( 3s+8\right ) \left ( s+4\right ) }-\frac{\left ( 2s+5\right ) }{\left ( s+2\right ) \left ( 3s+8\right ) }\\ & =\frac{2s+3}{s^{2}+6s+8}\\ & =\frac{2s+3}{\left ( s+4\right ) \left ( s+2\right ) } \end{align*}
Using Partial fractions, let \(\frac{2s+3}{\left ( s+4\right ) \left ( s+2\right ) }=\frac{A}{\left ( s+4\right ) }+\frac{B}{s+2}\), hence\begin{align*} A & =\lim _{s\rightarrow -4}\frac{\left ( 2s+3\right ) }{\left ( s+2\right ) }=\frac{\left ( -8+3\right ) }{\left ( -4+2\right ) }=\frac{5}{2}\\ B & =\lim _{s\rightarrow -2}\frac{\left ( 2s+3\right ) }{\left ( s+4\right ) }=\frac{\left ( -4+3\right ) }{\left ( -2+4\right ) }=\frac{-1}{2} \end{align*}
Or\begin{align*} Y\left ( s\right ) & =\frac{A}{\left ( s+4\right ) }+\frac{B}{s+2}\\ & =\frac{5}{2\left ( s+4\right ) }-\frac{1}{2\left ( s+2\right ) } \end{align*}
Then\[ y\left ( t\right ) =\frac{5}{2}e^{-4t}-\frac{1}{2}e^{-2t}\] Summary\begin{align*} z\left ( t\right ) & =\frac{1}{3}e^{-t}-\frac{10}{3}e^{-4t}+e^{-2t}\\ y\left ( t\right ) & =\frac{5}{2}e^{-4t}-\frac{1}{2}e^{-2t} \end{align*}
Solve for the currents in the circuit assuming currents and charges are initially zero and act that \(E\left ( t\right ) =2H\left ( t-4\right ) -H\left ( t-5\right ) \)
Answer:
For reference, these are the laws to remember: Ohm’s law \(V=Ri\) for voltage across resistor, \(V=Li^{\prime }\) for voltage across inductor, \(V=Q/C\) for voltage across capacitor, \(i=q^{\prime }\) for current capacitor relation, hence \(V=\frac{1}{c}\int _{0}^{t}i\left ( \tau \right ) d\tau \) for voltage across capacitor.
Applying Kirchoff’s voltage law to each loop gives\begin{align*} 5i_{1}+5\frac{d}{dt}i_{1}-5\frac{d}{dt}i_{2} & =E\left ( t\right ) \\ 5i_{2}+5\frac{d}{dt}i_{2}-5\frac{d}{dt}i_{1} & =0 \end{align*}
Taking Laplace transform, and using property \(\mathcal{L}\left ( H\left ( t-a\right ) \right ) =\frac{1}{s}e^{-as}\) gives (writing \(I_{1}\) to mean \(I_{1}\left ( s\right ) \) and \(I_{2}\) to mean \(I_{2}\left ( s\right ) \))\begin{align*} 5I_{1}+5\left ( sI_{1}-i_{1}\left ( 0\right ) \right ) -5\left ( sI_{2}-i_{2}\left ( 0\right ) \right ) & =\frac{2}{s}e^{-4s}-\frac{1}{s}e^{-5s}\\ 5I_{2}+5\left ( sI_{2}-i_{2}\left ( 0\right ) \right ) -5\left ( sI_{1}-i_{1}\left ( 0\right ) \right ) & =0 \end{align*}
Setting initial conditions, the above becomes\begin{align} I_{1}\left ( 5+5s\right ) -5sI_{2} & =\frac{2}{s}e^{-4s}-\frac{1}{s}e^{-5s}\tag{1}\\ I_{2}\left ( 5+5s\right ) -5sI_{1} & =0\nonumber \end{align}
Solving for \(I_{1}\left ( s\right ) \) from the second equation gives\[ \,I_{1}=\frac{I_{2}\left ( 5+5s\right ) }{5s}\] Substituting this into Eq. (1) gives\begin{align} \frac{I_{2}\left ( 5+5s\right ) }{5s}\left ( 5+5s\right ) -5sI_{2} & =\frac{2}{s}e^{-4s}-\frac{1}{s}e^{-5s}\nonumber \\ I_{2}\left ( 25s^{2}+50s+25\right ) -25s^{2}I_{2} & =10e^{-4s}-5e^{-5s}\nonumber \\ I_{2}\left ( 50s+25\right ) & =10e^{-4s}-5e^{-5s}\nonumber \\ I_{2} & =\frac{10e^{-4s}-5e^{-5s}}{50s+25}\nonumber \\ & =\frac{10}{25}\left ( \frac{e^{-4s}}{2s+1}\right ) -\frac{5}{25}\left ( \frac{e^{-5s}}{2s+1}\right ) \nonumber \\ & =\frac{10}{25}\frac{1}{2}\left ( \frac{e^{-4s}}{s+\frac{1}{2}}\right ) -\frac{5}{25}\frac{1}{2}\left ( \frac{e^{-5s}}{s+\frac{1}{2}}\right ) \nonumber \\ & =\frac{1}{5}\left ( \frac{e^{-4s}}{s+\frac{1}{2}}\right ) -\frac{1}{10}\left ( \frac{e^{-5s}}{s+\frac{1}{2}}\right ) \tag{2} \end{align}
To be able to use the property \(\mathcal{L}^{-1}\left ( \frac{e^{-as}}{\left ( s+b\right ) }\right ) =H\left ( t-a\right ) \mathcal{L}^{-1}\left ( \frac{1}{s+b}\right ) \) and the property that \(\mathcal{L}^{-1}\left ( \frac{1}{s+b}\right ) =e^{-bt}\) Hence the inverse Laplace transform is\begin{align*} i_{2}\left ( t\right ) & =\frac{1}{5}H\left ( t-4\right ) e^{\left ( -\frac{1}{2}\left ( t-4\right ) \right ) }-\frac{1}{10}H\left ( t-5\right ) e^{\left ( -\frac{1}{2}\left ( t-5\right ) \right ) }\\ & =\frac{1}{5}H\left ( t-4\right ) e^{\left ( -\frac{1}{2}t+2\right ) }-\frac{1}{10}H\left ( t-5\right ) e^{\left ( -\frac{1}{2}t+\frac{5}{2}\right ) } \end{align*}
Hence\[ i_{2}\left ( t\right ) =\frac{1}{5}e^{-\frac{t}{2}}\left ( e^{2}H\left ( t-4\right ) -\frac{1}{2}e^{\frac{5}{2}}H\left ( t-5\right ) \right ) \] Now \(I_{2}\) from Eq. (2) is substituted back into Eq. (1) to solve for \(I_{1}\). Hence Eq. (1) becomes\begin{align*} I_{1}\left ( 5+5s\right ) -5s\left ( \frac{1}{5}\left ( \frac{e^{-4s}}{s+\frac{1}{2}}\right ) -\frac{1}{10}\left ( \frac{e^{-5s}}{s+\frac{1}{2}}\right ) \right ) & =\frac{2}{s}e^{-4s}-\frac{1}{s}e^{-5s}\\ I_{1}\left ( 5+5s\right ) & =\frac{2}{s}e^{-4s}-\frac{1}{s}e^{-5s}+5s\left ( \frac{1}{5}\left ( \frac{e^{-4s}}{s+\frac{1}{2}}\right ) -\frac{1}{10}\left ( \frac{e^{-5s}}{s+\frac{1}{2}}\right ) \right ) \\ & =\frac{2}{s}e^{-4s}-\frac{1}{s}e^{-5s}+s\left ( \frac{e^{-4s}}{s+\frac{1}{2}}\right ) -\frac{1}{2}s\left ( \frac{e^{-5s}}{s+\frac{1}{2}}\right ) \\ & =\frac{\left ( s+\frac{1}{2}\right ) \frac{2}{s}e^{-4s}-\left ( s+\frac{1}{2}\right ) \frac{1}{s}e^{-5s}+se^{-4s}-\frac{1}{2}se^{-5s}}{s+\frac{1}{2}} \end{align*}
Hence\begin{align*} I_{1} & =\frac{\left ( s+\frac{1}{2}\right ) \frac{2}{s}e^{-4s}-\left ( s+\frac{1}{2}\right ) \frac{1}{s}e^{-5s}+se^{-4s}-\frac{1}{2}se^{-5s}}{5\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }\\ & =\frac{\left ( s+\frac{1}{2}\right ) \frac{2}{s}e^{-4s}}{5\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }-\frac{\left ( s+\frac{1}{2}\right ) \frac{1}{s}e^{-5s}}{5\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }+\frac{1}{5}\frac{se^{-4s}}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }-\frac{1}{10}\frac{se^{-5s}}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }\\ & =\frac{2}{5}\frac{e^{-4s}}{s\left ( 1+s\right ) }-\frac{1}{5}\frac{e^{-5s}}{s\left ( 1+s\right ) }+\frac{1}{5}\frac{se^{-4s}}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }-\frac{1}{10}\frac{se^{-5s}}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) } \end{align*}
The inverse Laplace transform of each term is now found. The following properties will be used\begin{align*} \mathcal{L}^{-1}\left ( \frac{e^{-as}}{\left ( s+b\right ) }\right ) & =H\left ( t-a\right ) \mathcal{L}^{-1}\left ( \frac{1}{s+b}\right ) =H\left ( t-a\right ) f\left ( t-a\right ) \\\mathcal{L}^{-1}\left ( \frac{1}{s+b}\right ) & =e^{-bt} \end{align*}
Finding partial fractions of \(\frac{1}{s\left ( 1+s\right ) }=\frac{A}{s}+\frac{B}{s+1}\), hence\begin{align*} A & =\lim _{s\rightarrow 0}\frac{1}{\left ( 1+s\right ) }=1\\ B & =\lim _{s\rightarrow -1}\frac{1}{s}=-1 \end{align*}
Hence\[ \frac{1}{s\left ( 1+s\right ) }=\frac{1}{s}-\frac{1}{s+1}\] Therefore, inverse Laplace of first term in Eq. (3) gives\begin{align*} \mathcal{L}^{-1}\left ( \frac{2}{5}\frac{e^{-4s}}{s\left ( 1+s\right ) }\right ) & =\frac{2}{5}H\left ( t-4\right ) \mathcal{L}^{-1}\left ( \frac{1}{s}-\frac{1}{s+1}\right ) \\ & =\frac{2}{5}H\left ( t-4\right ) \left ( H\left ( t-4\right ) -e^{-\left ( t-4\right ) }\right ) \\ & =\frac{2}{5}\left ( H\left ( t-4\right ) -H\left ( t-4\right ) e^{-\left ( t-4\right ) }\right ) \end{align*}
And for the second term in Eq. (3)\begin{align*} \mathcal{L}^{-1}\left ( \frac{1}{5}\frac{e^{-5s}}{s\left ( 1+s\right ) }\right ) & =\frac{1}{5}H\left ( t-5\right ) \mathcal{L}^{-1}\left ( \frac{1}{s}-\frac{1}{s+1}\right ) \\ & =\frac{1}{5}H\left ( t-5\right ) \left ( H\left ( t-4\right ) -e^{-\left ( t-5\right ) }\right ) \\ & =\frac{1}{5}\left ( H\left ( t-5\right ) -H\left ( t-5\right ) e^{-\left ( t-5\right ) }\right ) \end{align*}
For the third term, partial fractions of \(\frac{s}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }\) is needed. Let \[ \frac{s}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }=\frac{A}{\left ( 1+s\right ) }+\frac{B}{\left ( s+\frac{1}{2}\right ) }\]\begin{align*} A & =\lim _{s\rightarrow -1}\frac{s}{\left ( s+\frac{1}{2}\right ) }=\frac{-1}{-\frac{1}{2}}=2\\ B & =\lim _{s\rightarrow -\frac{1}{2}}\frac{s}{\left ( 1+s\right ) }=\frac{-\frac{1}{2}}{\left ( 1-\frac{1}{2}\right ) }=-1 \end{align*}
Hence \begin{align*} \mathcal{L}^{-1}\left ( \frac{1}{5}\frac{se^{-4s}}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }\right ) & =\frac{1}{5}H\left ( t-4\right ) \mathcal{L}^{-1}\left ( \frac{2}{\left ( 1+s\right ) }-\frac{1}{\left ( s+\frac{1}{2}\right ) }\right ) \\ & =\frac{1}{5}H\left ( t-4\right ) \left ( 2e^{-\left ( t-4\right ) }-e^{-\frac{1}{2}\left ( t-4\right ) }\right ) \end{align*}
For the last term in Eq. (3)\begin{align*} \mathcal{L}^{-1}\left ( \frac{1}{10}\frac{se^{-5s}}{\left ( 1+s\right ) \left ( s+\frac{1}{2}\right ) }\right ) & =\frac{1}{10}H\left ( t-5\right ) \mathcal{L}^{-1}\left ( \frac{2}{\left ( 1+s\right ) }-\frac{1}{\left ( s+\frac{1}{2}\right ) }\right ) \\ & =\frac{1}{10}H\left ( t-5\right ) \left ( 2e^{-\left ( t-5\right ) }-e^{-\frac{1}{2}\left ( t-5\right ) }\right ) \end{align*}
Putting all this together gives\begin{align*} i_{1}\left ( t\right ) & =\frac{2}{5}\left ( H\left ( t-4\right ) -H\left ( t-4\right ) e^{-\left ( t-4\right ) }\right ) \\ & -\frac{1}{5}\left ( H\left ( t-5\right ) -H\left ( t-5\right ) e^{-\left ( t-5\right ) }\right ) \\ & +\frac{1}{5}H\left ( t-4\right ) \left ( 2e^{-\left ( t-4\right ) }-e^{-\frac{1}{2}\left ( t-4\right ) }\right ) \\ & -\frac{1}{10}H\left ( t-5\right ) \left ( 2e^{-\left ( t-5\right ) }-e^{-\frac{1}{2}\left ( t-5\right ) }\right ) \end{align*}
This can be simplified to\begin{align*} i_{1}\left ( t\right ) & =\frac{2}{5}H\left ( t-4\right ) +H\left ( t-4\right ) \left ( -\frac{2}{5}e^{-\left ( t-4\right ) }+\frac{2}{5}e^{-\left ( t-4\right ) }-\frac{1}{5}e^{-\frac{1}{2}\left ( t-4\right ) }\right ) \\ & -\frac{1}{5}H\left ( t-5\right ) +H\left ( t-5\right ) \left ( \frac{1}{5}e^{-\left ( t-5\right ) }-\frac{1}{5}e^{-\left ( t-5\right ) }+\frac{1}{10}e^{-\frac{1}{2}\left ( t-5\right ) }\right ) \end{align*}
Hence, summary of final results:\begin{align*} i_{1}\left ( t\right ) & =\frac{2}{5}H\left ( t-4\right ) -\frac{1}{5}H\left ( t-5\right ) +\frac{1}{5}H\left ( t-4\right ) e^{-\frac{1}{2}\left ( t-4\right ) }+\frac{1}{10}H\left ( t-5\right ) e^{-\frac{1}{2}\left ( t-5\right ) }\\ i_{2}\left ( t\right ) & =\frac{1}{5}e^{-\frac{t}{2}}\left ( e^{2}H\left ( t-4\right ) -\frac{1}{2}e^{\frac{5}{2}}H\left ( t-5\right ) \right ) \end{align*}
Here is a plot of the solutions for \(t=0\cdots 8\sec \) showing the input \(E\left ( t\right ) \) and the currents \(i_{1}\left ( t\right ) \) and \(i_{2}\left ( t\right ) \).
Solution
Assuming the generalized coordinate for \(m_{1}\) is \(y_{1}\) and for \(m_{2}\) is \(y_{2}\) and these are measured when the system is relaxed, hence force due to weight is already accounted for. Assume positive is upwards. For mass \(m_{1}\) applying \(F=ma\) gives\[ -k_{1}y_{1}-k_{2}\left ( y_{1}-y_{2}\right ) +f_{1}\left ( t\right ) =m_{1}y_{1}^{\prime \prime }\left ( t\right ) \] and for mass \(m_{2}\)\[ -k_{3}y_{1}-k_{2}\left ( y_{1}-y_{2}\right ) =m_{2}y_{2}^{\prime \prime }\left ( t\right ) \] Or\begin{align} m_{1}y_{1}^{\prime \prime }\left ( t\right ) +y_{1}\left ( k_{1}+k_{2}\right ) -k_{2}y_{2} & =f_{1}\left ( t\right ) \tag{1}\\ m_{2}y_{2}^{\prime \prime }\left ( t\right ) -y_{2}\left ( k_{2}\right ) +y_{1}\left ( k_{3}+k_{2}\right ) & =0\nonumber \end{align}
Or, in matrix form\[\begin{pmatrix} m_{1} & 0\\ 0 & m_{2}\end{pmatrix}\begin{pmatrix} y_{1}^{\prime \prime }\\ y_{2}^{\prime \prime }\end{pmatrix} +\begin{pmatrix} k_{1}+k_{2} & -k_{2}\\ -k_{2} & k_{3}+k_{2}\end{pmatrix}\begin{pmatrix} y_{1}\\ y_{2}\end{pmatrix} =\begin{pmatrix} 1-H\left ( t-2\right ) \\ 0 \end{pmatrix} \] The stiffness matrix is symmetric and positive definite. OK. Now Laplace transform is applied to the above, using \(Y\) to mean \(Y\left ( s\right ) \), the Laplace transform of \(y\left ( t\right ) \), hence\[\begin{pmatrix} m_{1} & 0\\ 0 & m_{2}\end{pmatrix}\begin{pmatrix} s^{2}Y_{1}-sy_{1}\left ( 0\right ) -y_{1}^{\prime }\left ( 0\right ) \\ s^{2}Y_{2}-sy_{2}\left ( 0\right ) -y_{2}^{\prime }\left ( 0\right ) \end{pmatrix} +\begin{pmatrix} k_{1}+k_{2} & -k_{2}\\ -k_{2} & k_{3}+k_{2}\end{pmatrix}\begin{pmatrix} Y_{1}\\ Y_{2}\end{pmatrix} =\begin{pmatrix} \frac{1}{s}-\frac{1}{s}e^{-2s}\\ 0 \end{pmatrix} \] Applying IC, gives\[\begin{pmatrix} m_{1} & 0\\ 0 & m_{2}\end{pmatrix}\begin{pmatrix} s^{2}Y_{1}\\ s^{2}Y_{2}\end{pmatrix} +\begin{pmatrix} k_{1}+k_{2} & -k_{2}\\ -k_{2} & k_{3}+k_{2}\end{pmatrix}\begin{pmatrix} Y_{1}\\ Y_{2}\end{pmatrix} =\begin{pmatrix} \frac{1}{s}-\frac{1}{s}e^{-2s}\\ 0 \end{pmatrix} \] Hence\[\begin{pmatrix} m_{1}s^{2}+\left ( k_{1}+k_{2}\right ) & -k_{2}\\ -k_{2} & m_{2}s^{2}+\left ( k_{3}+k_{2}\right ) \end{pmatrix}\begin{pmatrix} Y_{1}\\ Y_{2}\end{pmatrix} =\begin{pmatrix} \frac{1}{s}-\frac{1}{s}e^{-2s}\\ 0 \end{pmatrix} \] To speed the typing, numerical values are now substituted for the symbolic values above (but it is best to delay this to the end otherwise so that same solution can be applied to different numerical values)\[\begin{pmatrix} s^{2}+8 & -2\\ -2 & s^{2}+5 \end{pmatrix}\begin{pmatrix} Y_{1}\\ Y_{2}\end{pmatrix} =\begin{pmatrix} \frac{1}{s}-\frac{1}{s}e^{-2s}\\ 0 \end{pmatrix} \] Hence\begin{align*} \begin{pmatrix} Y_{1}\\ Y_{2}\end{pmatrix} & =\begin{pmatrix} s^{2}+8 & -2\\ -2 & s^{2}+5 \end{pmatrix} ^{-1}\begin{pmatrix} \frac{1}{s}-\frac{1}{s}e^{-2s}\\ 0 \end{pmatrix} \\ & =\frac{1}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4}\begin{pmatrix} s^{2}+5 & 2\\ 2 & s^{2}+8 \end{pmatrix}\begin{pmatrix} \frac{1}{s}-\frac{1}{s}e^{-2s}\\ 0 \end{pmatrix} \end{align*}
Therefore \begin{align*} Y_{1} & =\frac{s^{2}+5}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4}\left ( \frac{1}{s}-\frac{1}{s}e^{-2s}\right ) \\ Y_{2} & =\frac{2}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4}\left ( \frac{1}{s}-\frac{1}{s}e^{-2s}\right ) \end{align*}
Or\begin{align*} Y_{1} & =\frac{1}{s}\frac{s^{2}+5}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4}-\frac{1}{s}e^{-2s}\frac{s^{2}+5}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4}\\ Y_{2} & =\frac{1}{s}\frac{2}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4}-\frac{1}{s}e^{-2s}\frac{2}{\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4} \end{align*}
But \(\left ( s^{2}+8\right ) \left ( s^{2}+5\right ) -4=\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) \), hence the above becomes (after writing \(s^{2}+5\) as \(\left ( s^{2}+4\right ) +1\))\begin{align} Y_{1} & =\frac{1}{s}\frac{\left ( s^{2}+4\right ) +1}{\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }-\frac{1}{s}e^{-2s}\frac{\left ( s^{2}+4\right ) +1}{\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\nonumber \\ Y_{2} & =\frac{1}{s}\frac{2}{\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }-\frac{1}{s}e^{-2s}\frac{2}{\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) } \tag{2} \end{align}
This can be written as\begin{align} Y_{1} & =\left ( \frac{1}{s\left ( s^{2}+9\right ) }\right ) +\left ( \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) -e^{-2s}\frac{1}{s\left ( s^{2}+9\right ) }-e^{-2s}\frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\nonumber \\ Y_{2} & =2\left ( \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) -2e^{-2s}\left ( \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) \tag{2} \end{align}
So there are only 3 common terms that needs to be inverse Laplace. These are \(\frac{1}{s\left ( s^{2}+9\right ) }\) and \(\frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\)
\[ \frac{1}{s\left ( s^{2}+9\right ) }=\frac{A}{s}+\frac{Bs+C}{s^{2}+9}\] Then\[ 1=A\left ( s^{2}+9\right ) +Bs^{2}+Cs \] Let \(s=0\) hence \(A=\frac{1}{9}\). Now, comparing \(s^{2}\) coefficients gives \(\frac{1}{9}+B=0\) or \(B=-\frac{1}{9}\). And finally comparing \(s\) coefficients shows that \(C=0\), therefore\[ \frac{1}{s\left ( s^{2}+9\right ) }=\frac{1}{9}\frac{1}{s}-\frac{1}{9}\frac{s}{s^{2}+9}\] Hence \[\mathcal{L}^{-1}\left ( \frac{1}{9}\frac{1}{s}-\frac{1}{9}\frac{s}{s^{2}+9}\right ) =\frac{1}{9}-\frac{1}{9}\cos 3t \] and\[\mathcal{L}^{-1}\left ( e^{-2s}\frac{1}{s\left ( s^{2}+9\right ) }\right ) =H\left ( t-2\right ) \left ( \frac{1}{9}H\left ( t-2\right ) -\frac{1}{9}\cos \left ( 3\left ( t-2\right ) \right ) \right ) \]
Now looking at the term \[ \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }=\frac{A}{s}+\frac{Bs+C}{s^{2}+4}+\frac{Ds+G}{s^{2}+9}\] Comparing coefficients gives \begin{align*} 1 & =A\left ( s^{4}+13s^{2}+36\right ) +\left ( Bs+C\right ) \left ( s^{3}+9s\right ) +\left ( Ds+G\right ) \left ( s^{3}+4s\right ) \\ 1 & =A\left ( s^{4}+13s^{2}+36\right ) +Bs^{4}+9Bs^{2}+Cs^{3}+9Cs+Ds^{4}+4Ds^{2}+Gs^{3}+4Gs \end{align*}
Hence\[ 1=s^{4}\left ( A+B+D\right ) +s^{3}\left ( C+G\right ) +s^{2}\left ( 13A+9B+4D\right ) +s\left ( 9C+4G\right ) +36A \] Therefore, \(A=\frac{1}{36}\) and\begin{align*} A+B+D & =0\\ C+G & =0\\ 13A+9B+4D & =0\\ 9C+4G & =0 \end{align*}
\(C=0\) and \(G=0\) since the second and the fourth equation do not have solution other than zero. Hence the above becomes\begin{align*} A+B+D & =0\\ 13A+9B+4D & =0 \end{align*}
From the first equation, \(B=-\frac{1}{36}-D\), hence from the second equation \(\frac{13}{36}+9\left ( -\frac{1}{36}-D\right ) +4D=0\) or \(\frac{4}{36}-5D=0\), hence \(D=\frac{4}{36\times 5}=\frac{1}{45}\). Therefore \(B=-\frac{1}{36}-D=-\frac{1}{36}-\frac{1}{45}=-\frac{1}{20}\)
Therefore\[ \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }=\frac{1}{36}\frac{1}{s}-\frac{1}{20}\frac{s}{s^{2}+4}+\frac{1}{45}\frac{s}{\left ( s^{2}+9\right ) }\] Now the inverse Laplace transform can be taken\[\mathcal{L}^{-1}\left ( \frac{1}{36}\frac{1}{s}-\frac{1}{20}\frac{s}{s^{2}+4}\allowbreak +\frac{1}{45}\frac{s}{\left ( s^{2}+9\right ) }\right ) =\frac{1}{36}-\frac{1}{20}\cos 2t+\frac{1}{45}\cos 3t \] Hence\[\mathcal{L}^{-1}\left ( e^{-2s}\frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) =H\left ( t-2\right ) \left ( \frac{1}{36}H\left ( t-2\right ) -\frac{1}{20}\cos 2\left ( t-2\right ) +\frac{1}{45}\cos 3\left ( t-2\right ) \right ) \]
Now that all terms have been inverse Laplace, the solution can be written down. From above,\begin{align} Y_{1} & =\left ( \frac{1}{s\left ( s^{2}+9\right ) }\right ) +\left ( \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) -e^{-2s}\frac{1}{s\left ( s^{2}+9\right ) }-e^{-2s}\frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\nonumber \\ Y_{2} & =2\left ( \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) -2e^{-2s}\left ( \frac{1}{s\left ( s^{2}+4\right ) \left ( s^{2}+9\right ) }\right ) \tag{2} \end{align}
Hence\begin{align*} y_{1}\left ( t\right ) & =\frac{1}{9}-\frac{1}{9}\cos 3t+\frac{1}{36}-\frac{1}{20}\cos 2t+\frac{1}{45}\cos 3t\\ & -H\left ( t-2\right ) \left ( \frac{1}{9}H\left ( t-2\right ) -\frac{1}{9}\cos \left ( 3\left ( t-2\right ) \right ) \right ) \\ & -H\left ( t-2\right ) \left ( \frac{1}{36}H\left ( t-2\right ) -\frac{1}{20}\cos 2\left ( t-2\right ) +\frac{1}{45}\cos 3\left ( t-2\right ) \right ) \end{align*}
Simplifying\[ y_{1}\left ( t\right ) =\frac{5}{36}-\frac{1}{20}\cos 2t-\frac{4}{45}\cos 3t-H\left ( t-2\right ) \left ( \frac{5}{36}-\frac{4}{45}\cos \left ( 3\left ( t-2\right ) \right ) -\frac{1}{20}\cos \left ( 2\left ( t-2\right ) \right ) \right ) \] and\begin{align*} y_{2}\left ( t\right ) & =\frac{2}{36}-\frac{2}{20}\cos 2t+\frac{2}{45}\cos 3t-2H\left ( t-2\right ) \left ( \frac{1}{36}H\left ( t-2\right ) -\frac{1}{20}\cos 2\left ( t-2\right ) +\frac{1}{45}\cos 3\left ( t-2\right ) \right ) \\ & =\frac{1}{18}-\frac{1}{10}\cos 2t+\frac{2}{45}\cos 3t-H\left ( t-2\right ) \left ( \frac{1}{18}-\frac{1}{10}\cos \left ( 2\left ( t-2\right ) \right ) +\frac{2}{45}\cos 3\left ( t-2\right ) \right ) \end{align*}
This is a plot of the solutions, with the input force
