Solution of the differential equation $y' = \sin(xy)$

Question

How can I solve the differential equation $y' = \sin(xy)$?

I have found it in an exercise of a book aboute ODE's, but its seems to me so much difficult I did not have idea how to try to solve it.

Thanks!

Answer 1

Let $u=xy$ ,

Then $y=\dfrac{u}{x}$

$\dfrac{dy}{dx}=\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}$

$\therefore\dfrac{1}{x}\dfrac{du}{dx}-\dfrac{u}{x^2}=\sin u$

$\dfrac{1}{x}\dfrac{du}{dx}=\dfrac{x^2\sin u+u}{x^2}$

$(x^2\sin u+u)\dfrac{dx}{du}=x$

Let $v=x^2$ ,

$\dfrac{dv}{du}=2x\dfrac{dx}{du}$

$\therefore\dfrac{(x^2\sin u+u)}{2x}\dfrac{dv}{du}=x$

$(x^2\sin u+u)\dfrac{dv}{du}=2x^2$

$(v\sin u+u)\dfrac{dv}{du}=2v$

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