Vtyjkyuk

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!

Let $u=xy$ ,

Then $y=dfracux$

$dfracdydx=dfrac1xdfracdudx-dfracux^2$

$thereforedfrac1xdfracdudx-dfracux^2=sin u$

$dfrac1xdfracdudx=dfracx^2sin u+ux^2$

$(x^2sin u+u)dfracdxdu=x$

Let $v=x^2$ ,

$dfracdvdu=2xdfracdxdu$

$thereforedfrac(x^2sin u+u)2xdfracdvdu=x$

$(x^2sin u+u)dfracdvdu=2x^2$

$(vsin u+u)dfracdvdu=2v$

Let $w=v+ucsc u$ ,

Then $v=w-ucsc u$

$dfracdvdu=dfracdwdu+(ucot u-1)csc u$

$therefore(sin u)wleft(dfracdwdu+(ucot u-1)csc uright)=2(w-ucsc u)$

$(sin u)wdfracdwdu+(ucot u-1)w=2w-2ucsc u$

$(sin u)wdfracdwdu=(3-ucot u)w-2ucsc u$

$wdfracdwdu=(3csc u-ucsc ucot u)w-2ucsc^2u$

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 $w=dfrac1z$ ,

Then $dfracdwdu=-dfrac1z^2dfracdzdu$

$therefore-dfrac1z^3dfracdzdu=dfrac3csc u-ucsc ucot uz-2ucsc^2u$

$dfracdzdu=2z^3ucsc^2u+(ucsc ucot u-3)z^2$

Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2