# Find solution of $\frac {d^2x}{dt^2}=-\sin x$

Problem: Find solution of $$\frac {d^2x}{dt^2}=-\sin x$$

Solution:Integrating both sides with respect to $t$

$$\frac {dx}{dt}=-t\sin x +c_1$$

Again integrating ,we get

$$x=\frac {-t^2 \sin x}{2} +tc_1+c_2$$

Am I doing right ?

Here answer is $x(t)=acost +bsint$ where $a,b$ are constants.

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Since $x=x(t)$ it is not necessarily the case that $\int f(x)dt$ is $f(x)t$. –  Sharkos Sep 17 '13 at 16:46
@rst It sounds like from your comments that you're supposed to use an appropriate approximation for $\sin x$ when $x$ is very small, and then solve the corresponding ODE. Try solving the ODE $\ddot{x} = - x$, noting that $\sin x$ is very close to $x$ for $x$ sufficiently small. –  A Blumenthal Sep 17 '13 at 18:04
@ABlumenthal,you are right.In that case ,we will get the ans.thanks a lot –  rst Sep 18 '13 at 3:13

Integrating $\sin x$ with respect to $t$ does not give $t\sin x$ since $x$ depends on $t$. To see why, note that the derivative with respect to $t$ of $t\sin x$ is $(t\sin x)'=\sin x+tx'\cos x$, not $\sin x$.

The equation to be solved implies that $2x'x''=-2x'\sin x$ hence $(x')^2=2\cos x+C$. From here things depend on $C$. For example if $C=2$, $x'=\pm2\cos(x/2)$ hence $$\pm t=\int\frac{\mathrm dx}{2\cos(x/2)}=\int\frac{\mathrm d(\sin(x/2))}{1-\sin^2(x/2)}=\frac12\log\left(\frac{1+\sin(x/2)}{1-\sin(x/2)}\right)+C,$$ which is equivalent to $$x=\pm2\arcsin\tanh(t+C).$$

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But answer is $x(t)=acost +bsint$ where $a,b$ are constants –  rst Sep 17 '13 at 17:09
In question $x$ denote angle of displacement.We need solution for sufficiently small angles of displacement –  rst Sep 17 '13 at 17:14
The function $x(t)=a\cos(t)+b\sin(t)$ does not satisfy the differential equation except if $a=b=0$. –  Did Sep 17 '13 at 19:25
Could you explain how you got $(x')^2=2\cos x+C$? It seems to me that you dropped $x'$ from $-2x'\sin x$. –  L. Xu Sep 17 '13 at 21:30
@L.Xu $2x'x''$ is the derivative of $(x')^2$ and $-2x'\sin x$ is the derivative of $2\cos x$. Hence... –  Did Sep 18 '13 at 6:42
Here, I assume the ODE as the below one: $$y''(x)+\sin(y(x))=0$$ Now, use the following way:
If $F(y,y',y'')=0$ is the OE which is free of variable $x$, so by setting $y'=u$ we have $$y'=u\frac{du}{dx}$$ according to the Chain Rule and so we have $$F\left(y,u,u\frac{du}{dx}\right)=0$$
$\ddot \smile \;+^1\quad$ I miss you, my friend. How's the internet situation? –  amWhy Sep 18 '13 at 12:55