Solution of $\frac{dy}{dx}=(1+x^2)(1+y^2

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Target Exam

CUET

Subject

-- Mathematics - Section B1

Chapter

Differential Equations

Question:

Solution of $\frac{dy}{dx}=(1+x^2)(1+y^2)$ is:

Options:

$\tan^{-1}y=x+\frac{x^3}{3}+c$

$\tan^{-1}y=x-\frac{x^3}{3}+c$

$\tan^{-1}y=x^2+\frac{x^3}{3}+c$

$\tan^{-1}y=x^2-\frac{x^3}{3}+c$

Correct Answer:

$\tan^{-1}y=x+\frac{x^3}{3}+c$

Explanation:

$\frac{dy}{dx}=(1+x^2)(1+y^2)$

so $\int\frac{1}{1+y^2}dy=\int(1+x^2)dx$ (Integrating both sides)

$⇒\tan^{-1}y=x+\frac{x^3}{3}+c$