(1+x2)(1+y2)+xydxdy=0
Integrating,
⇒∫(21+y22y)dy=−∫(x1+x21+x2)dx
⇒1+y2=−∫1+x2xdx−∫x21+x2xdx
Put 1+x2=t⇒x2=t2−1 to solve RHS 2nd integration,
⇒1+y2=−1+x2−∫(t2−1)ttdt
⇒1+y2+1+x2=−∫t2−11dt
⇒1+y2+1+x2=21ln[1+x2−11+x2+1]+C
JEE Main 2020 — Mathematics Calculus
The general solution of the differential equation 1+x2+y2+x2y2+xydxdy=0 (where C is a constant of integration)
Held on 6 Sept 2020 · Verified 6 Jul 2026.
1+y2+1+x2=21loge(1+x2+11+x2−1)+C
1+y2−1+x2=21loge(1+x2+11+x2−1)+C
1+y2+1+x2=21loge(1+x2−11+x2+1)+C
1+y2−1+x2=21loge(1+x2−11+x2+1)+C
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