Q.Solve: $(2D^2 + 5D + 2)y = e^{-\frac{1}{2}x}$

Find the roots of the auxiliary equation for the CF, then apply the "root coincides with the exponent" rule (multiply by x and divide by f′(m)) for the PI.

1. Given $(2D^2+5D+2)y=e^{-x/2}$, where $D\equiv\dfrac{d}{dx}$.
2. Auxiliary equation: $2m^2+5m+2=0$.
3. Solve: $m=\dfrac{-5\pm\sqrt{25-4(2)(2)}}{2(2)}=\dfrac{-5\pm\sqrt9}{4}=\dfrac{-5\pm3}{4}$, giving $m=-\tfrac12$ and $m=-2$.
4. Complementary function: $y_c=C_1e^{-x/2}+C_2e^{-2x}$.
5. For the particular integral with RHS $e^{ax}$ where $f(D)=2D^2+5D+2$: here $a=-\tfrac12$, and $f\left(-\tfrac12\right)=2\left(\tfrac14\right)+5\left(-\tfrac12\right)+2=\tfrac12-\tfrac52+2=0$ — so $a=-\tfrac12$ is exactly a root of the auxiliary equation (the usual $\dfrac{e^{ax}}{f(a)}$ rule fails, division by zero). …