Variation of Parameters

First, solve for the complementary solution for the homogeneous DE:

\(y^{''} - 5y^{'} + 6y = 0\)

We will try a solution \(y=e^{kt}\) with auxilliary equation \( k^2 -5 k + 6 = 0 \). The determinant is:

\(b^2 - 4ac = (-5)^2 - 4(1)(6) = 25 - 24 = 1 > 0\)

The solution has two real roots. Factoring the auxilliary equation yields:

\( (k-3)(k-2) = 0\)

The solutions are \(k=3,2\) so the general solution is:

\( y_c = c_1e^{3t} +c_2e^{2t}\)

Next we will solve the set of equations:

\( u_1^{'} y_1 + u_2^{'} y_2 = 0 \)

\( u_1^{'} y_1^{'} + u_2^{'} y_2^{'} = g(x) \)

In matrix form this is:

\( \left [ \begin{matrix} e^{3t} & e^{2t} \\ 3e^{3t} & 2e^{2t} \end{matrix} \right ] \left [ \begin{matrix} u_1^{'} \\ u_2^{'} \end{matrix} \right ] = \left [ \begin{matrix} 0 \\ 2e^{t} \end{matrix} \right ] \)

Use Cramer's Rule to solve the system of equations:

\( u_1^{'} = \cfrac{\begin{vmatrix} 0 & e^{2t} \\ 2e^{t} & 2e^{2t} \end{vmatrix}} {\begin{vmatrix} e^{3t} & e^{2t} \\ 3e^{3t} & 2e^{2t} \end{vmatrix}} = \cfrac{-2e^{3t}}{-e^{5t}} = 2e^{-2t}\)

\( u_2^{'} = \cfrac{\begin{vmatrix} e^{3t} & 0 \\ 3e^{3t} & 2e^{t} \end{vmatrix}} {\begin{vmatrix} e^{3t} & e^{2t} \\ 3e^{3t} & 2e^{2t} \end{vmatrix}} = \cfrac{2e^{4t}}{-e^{5t}} = -2e^{-t}\)

Now take the integral to solve for \(u_1, u_2\):

\( u_1 = \int{u_1^{'} dt} = \int{2e^{-2t} dt} = -e^{-2t}\)

\( u_2 = \int{u_2^{'} dt} = \int{-2e^{-t} dt} = 2e^{-t}\)

The general particular solution is :

\(y_p = u_1 y_1 + u_2 y_2\)

\(y_p = -e^{-2t}e^{3t} + 2e^{-t} e^{2t} = e^{t}\)

The general solution is:

\(y = y_c + y_p = c_1e^{3t} +c_2e^{2t} + e^{t}\)