Solve the differential equation by variation of parameters, subject to the initial conditions y(0) = 1, y'(0) = 0. 36y'' − y = xex/6

The homogeneous ODE[tex]36y''-y=0[/tex]has characteristic equation[tex]36r^2-1=0[/tex]with roots at [tex]r=\pm\dfrac16[/tex], and admits two linearly independent solutions,[tex]y_1=e^{x/6}[/tex][tex]y_2=e^{-x/6}[/tex]as the Wronskian is[tex]W(y_1,y_2)=\begin{vmatrix}e^{x/6}&e^{-x/6}\\\\\dfrac16e^{x/6}&-\dfrac16e^{-x/6}\end{vmatrix}=-\dfrac13\neq0[/tex]Variation of parameters has us looking for solutions of the form[tex]y_p=u_1y_1+u_2y_2[/tex]such that[tex]u_1=-\displaystyle\int\frac{y_2xe^{x/6}}{W(y_1,y_2)}\,\mathrm dx[/tex][tex]u_2=\displaystyle\int\frac{y_1xe^{x/6}}{W(y_1,y_2)}\,\mathrm dx[/tex]We have[tex]u_1=\displaystyle3\int x\,\mathrm dx=\dfrac{3x^2}2[/tex][tex]u_2=\displaystyle-3\int xe^{x/3}\,\mathrm dx=-9e^{x/3}(x-3)[/tex]and we get[tex]y_p=\dfrac{3x^2e^{x/6}}2-9e^{x/6}(x-3)[/tex]The general solution is[tex]y=y_c+y_p[/tex][tex]y=C_1e^{x/6}+C_2e^{-x/6}+\dfrac{3x^2e^{x/6}}2-9e^{x/6}(x-3)[/tex]The initial conditions tell us[tex]\begin{cases}1=C_1+C_2+27\\\\0=\dfrac{C_1}6-\dfrac{C_2}6-\dfrac92\end{cases}\implies C_1=\dfrac12,C_2=-\dfrac{53}2[/tex]so that the particular solution is[tex]y=\dfrac12e^{x/6}-\dfrac{53}2e^{-x/6}+\dfrac32x^2e^{x/6}-9e^{x/6}(x-3)[/tex]