Solve the differential equation dydx=1+x+y2+xy2, when y=0 and x=0.
Given that, dydx=1+x+y2+xy2
⇒dydx=(1+x)+y2(1+x)⇒dydx=(1+y2)(1+x)⇒dy1+y2=(1+x)dx
On integrating bothe sides, we get
tan−1y=x+x22+K ...(i)
when y=0 and x=0, then substituting these values in Eq. (i), we get
tan−1(0)=0+0+K⇒K=0⇒tan−1y=x+x22⇒y=tan(x+x22)