Solution of $\frac{dy}{dx}=(1+x^2)(1+y^2)$ is:
Answer & explanation
Correct answer: option 1
$\frac{dy}{dx}=(1+x^2)(1+y^2)$
so $\int\frac{1}{1+y^2}dy=\int(1+x^2)dx$ (Integrating both sides)
$⇒\tan^{-1}y=x+\frac{x^3}{3}+c$
Solution of $\frac{dy}{dx}=(1+x^2)(1+y^2)$ is:
Correct answer: option 1
$\frac{dy}{dx}=(1+x^2)(1+y^2)$
so $\int\frac{1}{1+y^2}dy=\int(1+x^2)dx$ (Integrating both sides)
$⇒\tan^{-1}y=x+\frac{x^3}{3}+c$