# Differential equation general solution based on discriminant

I need to solve the differential equation $y'' + y' = -2xe^{-x}$.

This gives $\lambda^2 + \lambda = 0$, $\Delta > 0$ so the general solution is $$y = c_1e^{\lambda_1x} + c_2e^{\lambda_2x}$$

But what if $\Delta$ on another differential equation is 0 or below 0? I can't find a documentation for this... My notes say that for $\Delta= 0$, $y = e^{\lambda x}(c_1 + c_2)$ and for $\Delta < 0$, $y = e^{ax} (c_1 \sin(bx) + c_2 \cos(bx))$ but I don't know if they are correct...

• Commented Jan 5, 2015 at 14:36
• If $\Delta=0$ then you should have $y=e^{\lambda x}(c_1+xc_2)$. But keep in mind that this is the general solution to the homogenous equation, your equations is not homogeneous and thus you need to modify the homogenous solution to solve your case. Commented Jan 5, 2015 at 14:37

To find the particular integral, you can do this: Write the equation as $(D^2+D)y=-2xe^{-x}$.
The particular integral is then $$y=\frac1{D^2+D}(-2xe^{-x})=\frac1{D(D+1)}(-2xe^{-x}) =\biggl(\dfrac1D-\dfrac1{D+1}\biggr)(-2xe^{-x}).$$ Now $\dfrac1D(-2xe^{-x})=e^{-x}\dfrac1{D-1}(-2x)=e^{-x}\dfrac1{1-D}(2x)=e^{-x}(1+D)(2x)=e^{-x}(2x+2)$.
Also $\dfrac1{D+1}(-2xe^{-x})=e^{-x}\dfrac1D(-2x)=e^{-x}(-x^2)$.
Thus $y=e^{-x}(2x+2)-e^{-x}(-x^2)=e^{-x}(x^2+2x+2)$.
The complementary function is $A+Be^{-x}$.
With $C=B+2$, the general solution is $y=A+e^{-x}(x^2+2x+C)$.