2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Have you heard of the method of variation of parameters? $\endgroup$ – Kavi Rama Murthy May 7 at 8:49
  • $\begingroup$ Yes, I heard about this method. But I miss it at uni and now don't understand it. @KaviRamaMurthy $\endgroup$ – Bambeil May 7 at 9:06
  • $\begingroup$ 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. $\endgroup$ – Kavi Rama Murthy May 7 at 9:09
  • $\begingroup$ See sosmath.com/diffeq/second/variation/variation.html $\endgroup$ – Kavi Rama Murthy May 7 at 9:12
2
$\begingroup$

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).

$\endgroup$
1
$\begingroup$

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 $u''-u=0$.

$\endgroup$
0
$\begingroup$

Hint:

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

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

$\endgroup$
0
$\begingroup$

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$

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.