The solution of the D.E 2x2ydydx=tan⁡x2y2−2xy2, given that y(1)=π2 is

we know ddxx2y2=x22ydy dx+y2 2xGiven equation 2x2ydydx+2xy2=tan(x2y2) ddxx2y2=tan⁡x2y2⇒∫cot⁡x2y2dx2y2=∫dx⇒log⁡sinx2y2=x+C Put x=1,y=π2 log sinπ2=1+C 0=1+C  then C=−1log⁡sinx2y2=x-1sin⁡x2y2=ex−1