# Differential equation $y'' - y + 2\sin(x)=0$

I need help and explanation with this differential equation. Actually I really don't know how to solve just this type of equations. So the problem: $$y'' - y + 2\sin(x)=0$$ In my opinion first of all we solve homogeneous equation $$y''-y=0$$ and the solution of this is $$y=c_1e^x+c_2e^{-x}$$. And after that to solve it with $$2\sin(x)$$. From this point I need help.

• Have you heard of the method of variation of parameters? – Kavi Rama Murthy May 7 at 8:49
• Yes, I heard about this method. But I miss it at uni and now don't understand it. @KaviRamaMurthy – Bambeil May 7 at 9:06
• The idea is to pretend that $c_1$ and $c_2$ are functions, plugin $y=c_1e^{x}+c_2e^{-x}$ in to the given equation and try to get a first order DE. – Kavi Rama Murthy May 7 at 9:09
• – Kavi Rama Murthy May 7 at 9:12

In this case there is a simple answer. You can just guess that $$\sin\, x$$ is a a particular solution so the general solution is $$c_1e^{x}+c_2e^{-x}+\sin\, x$$. In general you have to use the method of variation of parameters. (A search on Wikipedia will be useful).

Let $$u=y+k\sin x$$. Then $$u''=y''-k\sin x$$.

So, $$y''-y+2\sin x=u''+k\sin x-u+k\sin x+2\sin x=0$$

In particular, if $$k=-1$$, then $$u''-u=0$$.

Hint:

You can use that $$\sin x\propto e^{ix}-e^{-ix},$$ and for any function $$y=ae^{bx}$$,

$$y''-y=a(b^2-1)e^{bx}.$$

The given differential equation is

$$y'' - y + 2\sin\ x=0$$ $$\implies y'' - y =- 2\sin x \implies (D^2 -1)y=- 2\sin x$$

where $$D \equiv \frac{d}{dx}$$

I think you have an idea about how to find the Complementary Function (i.e., C.F.), (for your case, which is nothing but the solution of the homogeneous differential equation $$y'' - y =0$$).

Here C.F. is $$c_1e^{x}+c_2e^{-x}$$.

Now for the Particular Integral (i.e., P.I.) there are some general rules

If $$f(D)$$ can be expressed as $$\phi(D^2)$$ and $$\phi(-a^2)\neq 0$$, then

$$1.$$ $$\frac{1}{f(D)} \sin ax=\frac{1}{\phi(D^2)} \sin ax = \frac{1}{\phi(-a^2)} \sin ax$$

$$2.$$ $$\frac{1}{f(D)} \cos ax=\frac{1}{\phi(D^2)} \cos ax = \frac{1}{\phi(-a^2)} \cos ax$$

Note: If $$f(D)$$ can be expressed as $$\phi(D^2)=D^2+a^2$$, then $$\phi(-a^2)= 0$$.

$$1.$$ $$\frac{1}{f(D)} \sin ax =\frac{1}{\phi(D^2)} \sin ax=x\frac{1}{\phi'(D^2)} \sin ax= x \frac{1}{2D} \sin ax= -\frac{x}{2a} \cos ax$$.

$$2.$$ $$\frac{1}{f(D)} \cos ax =\frac{1}{\phi(D^2)} \cos ax=x\frac{1}{\phi'(D^2)} \cos ax= x \frac{1}{2D} \cos ax= \frac{x}{2a} \sin ax$$.

where $$\phi'(D^2)\equiv\frac{d}{dD}\phi(D^2)$$

So for your problem, P.I. is $$\frac{1}{D^2 -1} (-2\sin x)=-2[\frac{1}{-1^2 -1} \sin x]= \sin x$$

Hence the general solution is $$y=$$ C.F. $$+$$ P.I. $$= c_1e^{x}+c_2e^{-x}+\sin\, x$$