Solve the following differential equation
\( \frac{d^2y}{dx^2}+4 \frac{dy}{dx}+6y= 3x+8 \)The auxiliary equation is
\[ \begin{align} \lambda^2+4 \lambda +6 = 0 \\ \\ Then\ using\ the\ quadratic\ formula \\ \\ \lambda =\frac{-4 \pm \sqrt{4^2-4(1)(6)}}{2(1)} \\ \\ \lambda =-2 \pm \sqrt{2} \end{align} \]So our complementary function is
\( Ae^{-2x}(cos \sqrt{2}x+ \epsilon) \)Then because the RHS of the differential equation is linear (i.e. \( 3x +8\) ), we use \( y= \lambda x+ \mu \) as our trial solution in order to find the particular integral
\( So \quad \frac{dy}{dx} =\lambda \quad and \quad \frac {d^2y}{dx^2}=0 \)Substituting into the differential equation gives
\[ \begin{align} 0 + 4(\lambda) + 6(\lambda x + \mu) = 3x + 8 \\ \\ Then\ comparing\ x\ coefficients\,, \\ 6\lambda = 3 \\ \lambda = \frac{1}{2} \\ and\ comparing\ constants\\ 4(\frac{1}{2})+6 \mu = 8 \\ \mu = 1 \\ \end{align} \]Therefore our particular integral is \( \frac{1}{2}x+1 \) and our final solution (i.e. Complementary Function + Particular Integral) is
\( y = Ae^{-2x}(cos \sqrt{2}x+ \epsilon)+ \frac{1}{2}x+1 \)