How to solve this unfamiliar initial value problem? I haven't learned how to solve this problem because the differential equation is non-linear. What's my approach?
$$\frac{dx}{dt}=\sin(tx),\space x(0)=\pi \space\text{for}\space  0\le x\le 10$$
 A: Hint:
Let $u=tx$ ,
Then $x=\dfrac{u}{t}$ 
$\dfrac{dx}{dt}=\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}$
$\therefore\dfrac{1}{t}\dfrac{du}{dt}-\dfrac{u}{t^2}=\sin u$
$\dfrac{1}{t}\dfrac{du}{dt}=\dfrac{t^2\sin u+u}{t^2}$
$(t^2\sin u+u)\dfrac{dt}{du}=t$
Let $v=t^2$ ,
$\dfrac{dv}{du}=2t\dfrac{dt}{du}$
$\therefore\dfrac{(t^2\sin u+u)}{2t}\dfrac{dv}{du}=t$
$(t^2\sin u+u)\dfrac{dv}{du}=2t^2$
$(v\sin u+u)\dfrac{dv}{du}=2v$
Let $w=v+u\csc u$ ,
Then $v=w-u\csc u$
$\dfrac{dv}{du}=\dfrac{dw}{du}+(u\cot u-1)\csc u$
$\therefore(\sin u)w\left(\dfrac{dw}{du}+(u\cot u-1)\csc u\right)=2(w-u\csc u)$
$(\sin u)w\dfrac{dw}{du}+(u\cot u-1)w=2w-2u\csc u$
$(\sin u)w\dfrac{dw}{du}=(3-u\cot u)w-2u\csc u$
$w\dfrac{dw}{du}=(3\csc u-u\csc u\cot u)w-2u\csc^2u$
This belongs to an Abel equation of the second kind.
In fact all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{3\csc u-u\csc u\cot u}{z}-2u\csc^2u$
$\dfrac{dz}{du}=2z^3u\csc^2u+(u\csc u\cot u-3)z^2$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2
