(A) The given differential equation is: $\frac{dy}{dx} = (1+x^2)(1+y^2)$Separating the variables,we get: $\frac{dy}{1+y^2} = (1+x^2) dx$Integrating bo

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

(A) The given differential equation is: $\frac{dy}{dx} = (1+x^2)(1+y^2)$Separating the variables,we get: $\frac{dy}{1+y^2} = (1+x^2) dx$Integrating both sides: $\int \frac{dy}{1+y^2} = \int (1+x^2) dx$We know that $\int \frac{dy}{1+y^2} = \tan^{-1} y$ and $\int (1+x^2) dx = x + \frac{x^3}{3} + C$Thus,the general solution is: $\tan^{-1} y = x + \frac{x^3}{3} + C$

Class 12 Mathematics Differential Equations Variable separable type differential equations

Medium

Find the general solution of the differential equation: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

A
$\tan^{-1} y = x + \frac{x^3}{3} + C$
B
$\tan^{-1} y = x + x^3 + C$
C
$\tan^{-1} y = \frac{x^3}{3} + C$
D
$\tan^{-1} y = x^2 + \frac{x^3}{3} + C$

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Variable separable type differential equations

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Find the general solution of the differential equation: $\frac{dy}{dx} = (1+x^2)(1+y^2)$

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