So, we first need to convert this into a system. Here’s the change of variables,

\[\begin{align*}{x_1} & = y & \hspace{0.25in}{{x'}_1} & = y' = {x_2}\\ {x_2} & = y' & \hspace{0.25in}{{x'}_2} & = y'' = \frac{3}{2}y - \frac{5}{2}y' = \frac{3}{2}{x_1} - \frac{5}{2}{x_2}\end{align*}\]

The system is then,

\[\vec x' = \left( {\begin{array}{*{20}{c}}0&1\\{\frac{3}{2}}&{ - \frac{5}{2}}\end{array}} \right)\vec x\hspace{0.25in}\vec x\left( 0 \right) = \left( {\begin{array}{*{20}{c}}{ - 4}\\9\end{array}} \right)\]

where,

\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{{x_1}\left( t \right)}\\{{x_2}\left( t \right)}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}{y\left( t \right)}\\{y'\left( t \right)}\end{array}} \right)\]

Now we need to find the eigenvalues for the matrix.

\[\begin{align*}\det \left( {A - \lambda I} \right) & = \left| {\begin{array}{*{20}{c}}{ - \lambda }&1\\{\frac{3}{2}}&{ - \frac{5}{2} - \lambda }\end{array}} \right|\\ & = {\lambda ^2} + \frac{5}{2}\lambda - \frac{3}{2}\\ & = \frac{1}{2}\left( {\lambda + 3} \right)\left( {2\lambda - 1} \right)\hspace{0.25in}{\lambda _1} = - 3,\,\,\,{\lambda _2} = \frac{1}{2}\end{align*}\]

Now let’s find the eigenvectors.

\({\lambda _{\,1}} = - 3\) :

We’ll need to solve,

\[\left( {\begin{array}{*{20}{c}}3&1\\{\frac{3}{2}}&{\frac{1}{2}}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}3{\eta _1} + {\eta _2} = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{\eta _2} = - 3{\eta _1}\]

The eigenvector in this case is,

\[\vec \eta = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{ - 3{\eta _1}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 1 \right)}} = \left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right),\,\,\,\,\,\,\,{\eta _1} = 1\]

\({\lambda _{\,2}} = \frac{1}{2}\):

We’ll need to solve,

\[\left( {\begin{array}{*{20}{c}}{ - \frac{1}{2}}&1\\{\frac{3}{2}}&{ - 3}\end{array}} \right)\left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{{\eta _2}}\end{array}} \right) = \left( {\begin{array}{*{20}{c}}0\\0\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in} - \frac{1}{2}{\eta _1} + {\eta _2} = 0\hspace{0.25in} \Rightarrow \hspace{0.25in}{\eta _2} = \frac{1}{2}{\eta _1}\]

The eigenvector in this case is,

\[\vec \eta = \left( {\begin{array}{*{20}{c}}{{\eta _1}}\\{\frac{1}{2}{\eta _1}}\end{array}} \right)\hspace{0.25in} \Rightarrow \hspace{0.25in}{\vec \eta ^{\left( 2 \right)}} = \left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right),\,\,\,\,\,\,\,{\eta _1} = 2\]

The general solution is then,

\[\vec x\left( t \right) = {c_1}{{\bf{e}}^{ - 3t}}\left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right) + {c_2}{{\bf{e}}^{\frac{t}{2}}}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\]

Apply the initial condition.

\[\left( {\begin{array}{*{20}{c}}{ - 4}\\9\end{array}} \right) = \vec x\left( 0 \right) = {c_1}\left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right) + {c_2}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\]

This gives the system of equations that we can solve for the constants.

\[\left. {\begin{array}{*{20}{c}}{{c_1} + 2{c_2} = - 4}\\{ - 3{c_1} + {c_2} = 9}\end{array}} \right\}\hspace{0.25in} \Rightarrow \hspace{0.25in}{c_1} = - \frac{{22}}{7},\,\,\,{c_2} = - \frac{3}{7}\]

The actual solution to the system is then,

\[\vec x\left( t \right) = - \frac{{22}}{7}{{\bf{e}}^{ - 3t}}\left( {\begin{array}{*{20}{c}}1\\{ - 3}\end{array}} \right) - \frac{3}{7}{{\bf{e}}^{\frac{t}{2}}}\left( {\begin{array}{*{20}{c}}2\\1\end{array}} \right)\]

Now recalling that,

\[\vec x\left( t \right) = \left( {\begin{array}{*{20}{c}}{y\left( t \right)}\\{y'\left( t \right)}\end{array}} \right)\]

we can see that the solution to the original differential equation is just the top row of the solution to the matrix system. The solution to the original differential equation is then,

\[y\left( t \right) = - \frac{{22}}{7}{{\bf{e}}^{ - 3t}} - \frac{6}{7}{{\bf{e}}^{\frac{t}{2}}}\]

Notice that as a check, in this case, the bottom row should be the derivative of the top row.