# Finding a particular solution to a Second Order Differential Equation

I am attempting to do the following problem (8.6.26 in Mary Boas, mathematical methods in physical sciences 3rd Edition). It provides the following Auxiliary equation:

$$(D^2+1)y=8x \sin(x)$$ I know that a general solution can be written:$$y=y_c+y_p,$$ where $y_p$ is a particular solution, and $y_c$ is a characteristic equation. I have obtained the characteristic equation to be: $$y_c=c_1e^{-ix}+c_2e^{ix},$$ where $c_1$ and $c_2$ are arbitrary constants.

We are considering: $$y''+y=8x \sin(x)$$ to find the particular solution we can consider the imaginary part of: $$Y''+Y=8xe^{ix}$$ My problem is coming when looking at the particular solution later. I can see that the roots of the auxiliary equation are $\pm i,$ so it makes sense to assume a particular solution of form: $$Cxe^{ix}.$$ From this: $$y_p'=Cixe^{ix}+Ce^{ix},$$ $$y_p''=-Cxe^{ix}+2Ce^{ix}$$ plugging these back in i find $$C=-4xi$$ and $$y_p=-4x^2ie^{ix}$$ 'expanding this', and considering only the imaginary pary, I get: $$y_p=-4x^2 \cos(x)$$ However this is clearly wrong, as when I re-substitute it back in, i get: $$-8\cos(x)+8x\sin(x)=8x \sin(x),$$ I am assuming I am missing a term somewhere, but I have repeated the question at least five times now and cannot spot it. I am posting it here to see if someone can see where I have gone wrong. Thanks in advance

• You should have $$y_p''=-Cxe^{ix}+{2\color{red}{i}Ce^{ix}}$$ – John Doe Jan 7 '18 at 20:27

The mistake happened when you calculated $y_p''$. It should be $$y_p''=-Cxe^{ix}+2iCe^{ix}$$
Also, the current $y_p$ provides a solution to $(D^2+1)y=8\sin(x)$, so you are looking for a particular solution of one order higher for $(D^2+1)y=8x\sin(x)$: $$y_p=Cxe^{ix}+Dx^2e^{ix}$$
Solving this correctly should give us that $$Im(y_p)=-2x^2\cos(x)+2x\sin{x}$$
As I mentioned in a comment, you computed $y_p''$ wrong. You should have $y_p''=-Cxe^{ix}+2iCe^{ix}$, which would give $$2iC e^{ix}=8xe^{ix}$$ Once you reached this point, you declared that $C=-4xi$. You cannot do this, since you assumed $C$ is a constant, and so it cannot depend on $x$. So clearly, we have not got a particular solution of this form. If we add a term though, we can get a solution. Take $$y_p=Cxe^{ix}+Dx^2e^{ix}$$ Then $$y_p''+y_p=2iCe^{ix}+2De^{ix}+4iDxe^{ix}=8xe^{ix}$$This gives that $D=-2i$, and $C=2$. So $$y_p=2xe^{ix}-2ix^2e^{ix}$$ The imaginary part of this is $$2x\sin x-2x^2\cos x$$(in agreement with the other answer)