3.2.2.1.4 Example 4 \(y^{\prime \prime }+2y^{\prime }+10y=\delta \left ( t\right ) \)

\begin{align*} y^{\prime \prime }+2y^{\prime }+10y & =\delta \left ( t\right ) \\ y\left ( 0\right ) & =0\\ y^{\prime }\left ( 0\right ) & =0 \end{align*}

Taking Laplace transform gives

\[ \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) +2\left ( sY-y\left ( 0\right ) \right ) +10Y=1 \]

Since given initial conditions then the above becomes

\begin{align*} s^{2}Y+2sY+10Y & =1\\ Y & =\frac {1}{s^{2}+2s+10}\\ & =\frac {1}{\left ( s+2\right ) \left ( s+5\right ) }\end{align*}

Taking inverse Laplace transform gives

\begin{align*} y & =\frac {1}{6}ie^{\left ( -1-3i\right ) t}-\frac {1}{6}ie^{\left ( -1+3i\right ) t}\\ & =\frac {1}{6}ie^{-t}e^{-3it}-\frac {1}{6}ie^{-t}e^{3it}\\ & =\frac {1}{6}ie^{-t}\left ( e^{-3it}-e^{3it}\right ) \\ & =\frac {1}{6}ie^{-t}\left ( \cos 3t-i\sin 3t-\left ( \cos 3t+i\sin 3t\right ) \right ) \\ & =\frac {1}{6}ie^{-t}\left ( -i\sin 3t-i\sin 3t\right ) \\ & =\frac {1}{6}ie^{-t}\left ( -2i\sin 3t\right ) \\ & =\frac {1}{3}e^{-t}\sin 3t \end{align*}

Which is the same as

\[ y=\left ( \frac {1}{3}e^{-t}\sin \left ( 3t\right ) \right ) U\left ( t\right ) \]

Where \(U\left ( t\right ) \) is Heaviside function which is one for \(t>0\). Note that it seems one should not give IC at same point of application of \(\delta \left ( t\right ) \) as in this problem. So this problem might be ill posed. Need to look more into this.