Solve the following differential equation
\( \frac{d^2y}{dx^2}+6 \frac{dy}{dx}+9y= e^{2x} \)The auxiliary equation is
\[ \begin{align} \lambda^2+6 \lambda +9 & = 0 \\ (\lambda+3)(\lambda +3) & = 0 \\ \therefore \quad \lambda =-3 \,, or\ \lambda & = -3 \\ \end{align} \]So our complementary function is
\[ \begin{align} e^{-3x}(A+Bx) \\ \end{align} \]Then because the RHS of the differential equation is \( e^{2x} \), we use \(y= \lambda e^{2x} \) as our trial solution in order to find the particular integral.
\[ \begin{align} So \quad \frac{dy}{dx}=2\lambda e^{2x} \quad and \quad \frac {d^2y}{dx^2}=4 \lambda e^{2x} \\ \end{align} \]Substituting into the differential equation gives
\[ \begin{align} 4 \lambda e^{2x}+6(2 \lambda e^{2x})+9 \lambda e^{2x} = e^{2x} \\ \implies \lambda= \frac{1}{25} \\ \end{align} \]Therefore our particular integral is \(\frac{1}{25}e^{2x} \) and our final solution (i.e. Complementary Function + Particular Integral) is \[ \begin{align} y = e^{-3x}(A+Bx)+ \frac{1}{25}e^{2x} \\ \end{align} \]