Solve the following differential equation
\( \frac{d^2y}{dx^2}-7 \frac{dy}{dx}+12y= 5 \)The auxiliary equation is
\[ \begin{align} \lambda^2-7 \lambda +12 & = 0 \\ (\lambda-4)(\lambda - 3) & = 0 \\ \therefore \quad \lambda =4 \,, or\ \lambda & = 3 \\ \end{align} \]So our complementary function is
\( Ae^{4x}+Be^{3x} \)Then because the RHS of the differential equation is a constant we use \(y= \lambda\) as our trial solution in order to find the particular integral
\(So \quad \frac{dy}{dx}=0 \quad and \quad \frac {d^2y}{dx^2}=0 \)Substituting into the differential equation gives
\[ \begin{align} 0-7(0)+12 \lambda = 5 \\ \implies \lambda= \frac{5}{12} \end{align} \]Therefore our particular integral is \( \frac{5}{12} \) and our final solution (i.e. Complementary Function + Particular Integral) is
\( y = Ae^{4x}+Be^{3x}+ \frac{5}{12}\)