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My attempt: The null-solution is $A\cos{2x}+B\sin{2x}.$ Let's start by rewriting RHS by

$$(1+\sin{x})^2=1+2\sin{x}+\sin^2{x}=\frac{3}{2}+2\sin{x}-\frac{1}{2}\cos{x}.$$ So we can now look at too differential equations:

$$\left\{ \begin{array}{rcr} y_1''+4y_1 & = & \frac{3}{2}+2\sin{x} \quad \quad \quad \quad (1)\\ y_2''+4y_2 & = & -\frac{1}{2}\cos{x} \quad \quad \quad \quad (2)\\ \end{array} \right.$$


For (1) we can assume that the particular solution is of the form $y_1=a+b\cos{x}+c\sin{x}$ and then $y_1'=-b\sin{x}+c\cos{x}$ and $y_1''=-b\cos{x}-c\sin{x}$ so

$$y_1''+4y_1= -b\cos{x}-c\sin{x} + 4(a+b\cos{x}+c\sin{x})=4a+3b\cos{x}+3c\sin{x}.$$

Identifying and solving coefficients gives $(a,b,c)=(\frac{3}{8},0,\frac{2}{3}),$ so I have that $y_{1p}=\frac{3}{8}+\frac{2}{3}\sin{x}.$


For (2) we can assume the same form for the particular solution. I get that

\begin{array}{lcl} y_2 & = & a+b\cos{x}+c\sin{x} \\ y_2' & = & -b\sin{x}+c\cos{x} \\ y_2''& = & -b\cos{x}-c\sin{x} \end{array}

Substituting in equation (2) gives $3a\cos{x}+3b\sin{x}=-\frac{1}{2}\cos{x}$, thus $(a,b)=(-\frac{1}{6},0)$ so $y_{2p}=-\frac{1}{6}\cos{x}.$


According to the law of super position I have that $$y_p=y_{1p}+y_{2p}=\frac{3}{8}+\frac{2}{3}\sin{x}-\frac{1}{6}\cos{x},$$

which finally gives

$$y(x)=y_h+y_p=A\cos{2x}+B\sin{2x}+\frac{2}{3}\sin{x}-\frac{1}{6}\cos{x}+\frac{3}{8}$$

Correct answer: $$y(x)=A\cos{2x}+B\sin{2x}+\frac{2}{3}\sin{x}-\frac{x}{8}\sin{2x}+\frac{3}{8}.$$

One term difference...why? Please note that I'm not interested in other solutions, I just want to know why my $y_{p2}$ is incorrect.

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  • $\begingroup$ In the first identity, you wrote $\cos x$ instead of $\cos2x$. $\endgroup$ – Yves Daoust Nov 19 '17 at 21:16
  • $\begingroup$ True. Thats an error there. But wouldn't that just make my $y_{2p}$ to $-\frac{1}{6}\cos{2x}$ instead? $\endgroup$ – Parseval Nov 19 '17 at 21:19
  • $\begingroup$ Hehe, not at all. Try it. $\endgroup$ – Yves Daoust Nov 19 '17 at 21:19
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For a particular solution, you can solve for every term independently

$$y''+4y=1\to y=\frac 14.$$ $$y''+4y=\sin x\to y=\frac23\sin x$$ (by indeterminate coefficients).

Then $$y''+4y=\cos2x\to ???$$ (indeterminate coefficients don't seem to work).

It turns out that this RHS can be expressed as an instance of the homogenous solution, so that you need another ansatz.

With $x(a\cos2x+\sin2x)$, you have

$$x(-4a\cos2x-4b\sin2x+4a\cos2x+4b\sin2x)+2(2a\sin2x-2b\cos2x)=\cos2x,$$ giving

$$y''+4y=\cos2x\to \frac{x\sin 2x}4.$$

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  • $\begingroup$ @ Yves Daoust: You say that the method of indeterminate coefficients don't seem to work. How does it seem so? To me, nothing indicates that it doesn't work, except for an answer different from that of the book. $\endgroup$ – Parseval Nov 21 '17 at 8:03

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