3.2.2.1.2 Example 2 \(y^{\prime \prime }-2y^{\prime }-3y=0\)

\begin{align*} y^{\prime \prime }-2y^{\prime }-3y & =0\\ y\left ( 4\right ) & =-3\\ y^{\prime }\left ( 4\right ) & =-17 \end{align*}

Taking Laplace transform gives

\[ \left ( s^{2}Y-sy\left ( 0\right ) -y^{\prime }\left ( 0\right ) \right ) -2\left ( sY-y\left ( 0\right ) \right ) -3Y=0 \]

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 ) -3Y & =0\\ Y\left ( s^{2}-2s-3\right ) -sc_{1}-c_{2}+2c_{1} & =0\\ Y & =\frac {sc_{1}+c_{2}-2c_{1}}{s^{2}-2s-3}\end{align*}

Taking inverse Laplace gives

\begin{equation} y\left ( t\right ) =\frac {1}{4}e^{-t}\left ( c_{2}\left ( e^{4t}-1\right ) +c_{1}\left ( 3+e^{4t}\right ) \right ) \tag {1}\end{equation}

Hence

\begin{equation} y^{\prime }\left ( t\right ) =\frac {1}{4}e^{-t}(4c_{1}e^{-4t}+4c_{2}e^{4t})-\frac {1}{4}e^{-t}(c_{2}(-1+e^{4t})+c_{1}(3+e^{4t})) \tag {2}\end{equation}

At \(t=4\) then (1,2) become

\begin{align*} -3 & =\frac {1}{4}e^{-4}\left ( c_{2}\left ( e^{16}-1\right ) +c_{1}\left ( 3+e^{16}\right ) \right ) \\ -17 & =\frac {1}{4}e^{-4}(4c_{1}e^{-16}+4c_{2}e^{16})-\frac {1}{4}e^{-4}(c_{2}(-1+e^{16})+c_{1}(3+e^{16})) \end{align*}

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

\begin{align*} c_{1} & =\frac {-5+2e^{16}}{e^{12}}\\ c_{2} & =\frac {-15-2e^{16}}{e^{12}}\end{align*}

Hence the solution (1) becomes

\begin{align*} y\left ( t\right ) & =\frac {1}{4}e^{-t}\left ( \frac {-15-2e^{16}}{e^{12}}\left ( e^{4t}-1\right ) +\frac {-5+2e^{16}}{e^{12}}\left ( 3+e^{4t}\right ) \right ) \\ & =-e^{3t}\left ( 5e^{-12}-2e^{4}e^{-4t}\right ) \\ & =-5e^{3t-12}+2e^{4-t}\end{align*}