The solution of primitive integral equation x2+y2dy=xydx is y=y(x) . If y(1)=1 and yx0=e then x0 is
2(e2−1)
2(e2+1)
3 e
e2+12
Put y=Vx ⇒ dydx=V+xdvdxfrom given dydx=xyx2+y2V+xdvdx=Vx2x21+V2=V1+V2xdvdx=V−V−V31+V2=−V31+V21+V2V3dv=-dxx→∫1V3+V2V3dv=−∫dxx→−12V2+logV=−logx+C→−x22y2+logy−logx=−logx+C
→logy−x22y2=CPasses through (1,1)_
log1-12=C, C=−12⇒logy−x22e2=−12 Put y=e⇒loge −x22e2=−121+12=x22e2⇒x22e2=32⇒x2=3e2⇒x02=3e2⇒x0=3e