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...

  • $\begingroup$ Here's a MathJax tutorial :) $\endgroup$ – Shaun Jan 5 '15 at 14:36
  • $\begingroup$ 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. $\endgroup$ – Wintermute Jan 5 '15 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 $y=A+e^{-x}(x^2+2x+C)$.


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