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The solution of the differential equation $\frac{dy}{dx}= (1 + x^2)(1 + y^2)$ is (Here C is an arbitrary constant) |
$\tan^{-1}y+x+\frac{x^3}{3}+=C$ $\tan^{-1}y-x-\frac{x^3}{3}+=C$ $\tan^{-1}x-y-\frac{x^3}{3}+=C$ $\tan^{-1}x+y+\frac{x^3}{3}+=C$ |
$\tan^{-1}y-x-\frac{x^3}{3}+=C$ |
The correct answer is Option (2) → $\tan^{-1}y-x-\frac{x^3}{3}+=C$ $\frac{dy}{dx}=(1+x^{2})(1+y^{2})$ $\Rightarrow \frac{dy}{1+y^{2}}=(1+x^{2})\,dx$ $\Rightarrow \tan^{-1}y= x+\frac{x^{3}}{3}+C$ $\displaystyle y=\tan\!\left(x+\frac{x^{3}}{3}+C\right)$ |
