We have
2y exydx+(y−2x exy)dy=0
⇒dxdy=2x exy−y2y exy ...............(1)
Here, the given differential equation is a homogeneous differential equation. As the right hand side of equation (1) is expressed as function of xy.
Therefore, we put x=vy
∴dxdy=v+ydvdy
substituting x=vy in the given differential equation we will get,
ydvdy=2vev−12ev−v
⇒ydvdy=−12ev
⇒2ev dv=−1ydy,y≠0
⇒2∫ev dv=−∫1ydy
⇒2ev=−log|y|+C
⇒2exy+log|y|=C ...........(2)
It is given that x=0 when y=1,
So putting x=0,y=1 in equation (2), we get
2e0+log 1=C
⇒C=2
Putting C=2 in equation (2), we get the required particular solution
2 exy+log y=2