Second order differential, getting 0 on the left side

Question

If I have $$y''+y'-6y = e^{2x}$$

and I choose $$Ce^{2x}$$ as my guess

then when computing the left side I get zero, what do I do to prevent this?

Answer 1

The problem is that the "associated homogeneous equation", y''+ y'- 6y= 0 has characteristic equation $r^2+ r- 6= (r+ 3)(r- 2)= 0$ so that r= 2 and -3. That is, the general solution to the associated homogeneous equation is $Ae^{2x}+ Be^{-3x}$. Since $e^{2x}$ satisfies the associated homogeneous equation, $Ce^{2x}$ cannot give $e^{2x}$ for any C. The standard thing to try instead is to use $y= Cxe^{2x}$. That way, using the product rule to differentiate you will have a part where you have differentiated the "$e^{2x}$", which will give 0, and a part where you have differentiated "x".

With $y= Cxe^{2x}$, $y'= Ce^{2x}+ 2Cxe^{2x}$ and $y''= 2Ce^{2x}+ 2Ce^{2x}+ 4xe^{2x}= 4Ce^{2x}+ 4Cxe^{2x}$ so that $y''+ y'- 6y= 4Ce^{2x}+ 4Cxe^{2x}+ Ce^{2x}+ 2Cxe^{2x}- 6Cxe^{2x}= 5Ce^{2x}= e^{2x}$. Notice that the "$Cxe^{2x}$ terms cancelled. $Cxe^{2x}$ satisfies the equation if C= 1/5.

What textbook are you using? Every Differential Equations text that I have seen covers this in detail.