3.2.2.1.3 Example 3 \(y^{\prime \prime }+2y^{\prime }+5y=50t-100\)

\begin{align*} y^{\prime \prime }+2y^{\prime }+5y & =50t-100\\ y\left ( 2\right ) & =-4\\ y^{\prime }\left ( 2\right ) & =14 \end{align*}

Taking Laplace transform gives

\[ \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +5Y=\frac {50}{s^{2}}-\frac {100}{s}\]

Since given initial conditions are not at \(t=0\), then let \(y\left ( 0\right ) =c_{1},y^{\prime }\left ( 0\right ) =c_{2}\) and the above becomes

\begin{align*} \left ( s^{2}Y-sc_{1}-c_{2}\right ) +2\left ( sY-c_{1}\right ) +5Y & =\frac {50}{s^{2}}-\frac {100}{s}\\ Y\left ( s^{2}+2s+5\right ) -sc_{1}-c_{2}-2c_{1} & =\frac {50}{s^{2}}-\frac {100}{s}\\ Y & =\frac {sc_{1}+c_{2}+2c_{1}+\frac {50}{s^{2}}-\frac {100}{s}}{s^{2}+2s+5}\end{align*}

Taking inverse Laplace gives

\begin{equation} y\left ( t\right ) =-24+10t+\left ( 24+c_{1}\right ) e^{-t}\cos \left ( 2t\right ) +\left ( 14+c_{1}+c_{2}\right ) e^{-t}\cos t\sin t \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }\left ( t\right ) =e^{-t}\left ( 10e^{t}+\left ( c_{2}-10\right ) \cos \left ( 2t\right ) -\left ( 110+5c_{1}+c_{2}\right ) \cos t\sin t\right ) \tag {2}\end{equation}

At \(t=2\) then (1,2) become

\begin{align*} -4 & =-24+20+\left ( 24+c_{1}\right ) e^{-2}\cos \left ( 4\right ) +\left ( 14+c_{1}+c_{2}\right ) e^{-2}\cos 2\sin 2\\ 14 & =e^{-2}\left ( 10e^{2}+\left ( c_{2}-10\right ) \cos \left ( 4\right ) -\left ( 110+5c_{1}+c_{2}\right ) \cos 2\sin 2\right ) \end{align*}

Solving the above for \(c_{1},c_{2}\) gives

\begin{align*} c_{1} & =-2\left ( 12+e^{2}\sin 4\right ) \\ c_{2} & =2\left ( 5+e^{2}\left ( 2\cos 4+\sin 4\right ) \right ) \end{align*}

Hence the solution (1) becomes

\[ y\left ( t\right ) =-24+10t+\left ( 24-2\left ( 12+e^{2}\sin 4\right ) \right ) e^{-t}\cos \left ( 2t\right ) +\left ( 14-2\left ( 12+e^{2}\sin 4\right ) +2\left ( 5+e^{2}\left ( 2\cos 4+\sin 4\right ) \right ) \right ) e^{-t}\cos t\sin t \]

Which simplifies to

\[ y\left ( t\right ) =-24+10t-2e^{2-t}\sin \left ( 4-2t\right ) \]