differential-equation with trigonometric functions Is there a "simple" way to solve a differential-equation like the following example $f'(t)=\sin(f(t))+t^2$?
I don't know how to approach this question, can someone help me? 
 A: No simple way.
Maple finds no closed form solution.  I doubt very much that there is one.
Of course you can have series solutions and numerical solutions.
For example, with initial condition $f(0)=0$ we get a series solution
$$ f \left( t \right) ={\frac{1}{3}}{t}^{3}+{\frac{1}{12}}{t}^{4}+{\frac
{1}{60}}{t}^{5}+{\frac{1}{360}}{t}^{6}+{\frac{1}{2520}}{t}^{7}+{\frac{
1}{20160}}{t}^{8}+{\frac{1}{181440}}{t}^{9}-{\frac{373}{604800}}{t}^{
10}+\ldots
$$
I suspect "this question" did not actually ask you to solve the differential equation...
A: Hint:
Follow the method in http://science.fire.ustc.edu.cn/download/download1/book%5Cmathematics%5CHandbook%20of%20Exact%20Solutions%20for%20Ordinary%20Differential%20EquationsSecond%20Edition%5Cc2972_fm.pdf#page=223:
Let $u=\tan\dfrac{f}{2}$ ,
Then $\dfrac{du}{dt}=\dfrac{t^2u^2}{2}+u+\dfrac{t^2}{2}$
Let $u=-\dfrac{2}{t^2v}\dfrac{dv}{dt}$ ,
Then $\dfrac{du}{dt}=-\dfrac{2}{t^2v}\dfrac{d^2v}{dt^2}+\dfrac{4}{t^3v}\dfrac{dv}{dt}+\dfrac{2}{t^2v^2}\left(\dfrac{dv}{dt}\right)^2$
$\therefore-\dfrac{2}{t^2v}\dfrac{d^2v}{dt^2}+\dfrac{4}{t^3v}\dfrac{dv}{dt}+\dfrac{2}{t^2v^2}\left(\dfrac{dv}{dt}\right)^2=\dfrac{2}{t^2v^2}\left(\dfrac{dv}{dt}\right)^2-\dfrac{2}{t^2v}\dfrac{dv}{dt}+\dfrac{t^2}{2}$
$\dfrac{2}{t^2v}\dfrac{d^2v}{dt^2}-\dfrac{2}{t^2v}\dfrac{dv}{dt}-\dfrac{4}{t^3v}\dfrac{dv}{dt}+\dfrac{t^2}{2}=0$
$4t\dfrac{d^2v}{dt^2}-4(t+2)\dfrac{dv}{dt}+t^5v=0$
