The general solution of the differential equa | vedclass

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Question

(B) To find the general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$,we use the method of separation of variables.Step

Vedclass · Accepted Answer

$	an^{-1} y = 	an^{-1} x + c$

Vedclass · Answer

(B) To find the general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$,we use the method of separation of variables.Step $1$: Separate the variables $x$ and $y$:$\frac{dy}{1+y^2} = \frac{dx}{1+x^2}$Step $2$: Integrate both sides:$\int \frac{dy}{1+y^2} = \int \frac{dx}{1+x^2}$Step $3$: Apply the standard integration formula $\int \frac{du}{1+u^2} = 	an^{-1} u + c$:$	an^{-1} y = 	an^{-1} x + c$Thus,the general solution is $	an^{-1} y = 	an^{-1} x + c$.

The general solution of the differential equation $\frac{dy}{dx} = \frac{1+y^2}{1+x^2}$ is . . . . . . .

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