dxdy−y=(ex)y2 ...(1)
taking y=v1
dxdy=−v21.dxdv
From (1)
−v21.dxdv−v1=ex.v21
dxdv+v=−ex...(2)
The integrating factor =e∫1dx=ex
Thus, the solution of (2) is
v.ex=∫(−ex).exdx+c
v.ex=−∫e2xdx+c
v.ex=−2e2x+c
yex=−2e2x+c
y1=2−ex+2c.e−x
y=−ex+2c.e−x2
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