The general solution of the differential equa | vedclass

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Question

(A) Given differential equation: $y dx + (1 + x^2) 	an^{-1} x dy = 0$Rearranging the terms to separate variables:$y dx = -(1 + x^2) 	an^{-1} x dy$$\

Vedclass · Accepted Answer

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

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(A) Given differential equation: $y dx + (1 + x^2) 	an^{-1} x dy = 0$Rearranging the terms to separate variables:$y dx = -(1 + x^2) 	an^{-1} x dy$$\frac{dx}{(1 + x^2) 	an^{-1} x} = -\frac{dy}{y}$Integrating both sides:$\int \frac{dx}{(1 + x^2) 	an^{-1} x} = -\int \frac{dy}{y}$Let $u = 	an^{-1} x$,then $du = \frac{1}{1 + x^2} dx$. The integral becomes:$\int \frac{du}{u} = -\int \frac{dy}{y}$$\ln|u| = -\ln|y| + \ln|c|$$\ln|	an^{-1} x| + \ln|y| = \ln|c|$$\ln|y 	an^{-1} x| = \ln|c|$Taking the exponential of both sides:$y 	an^{-1} x = c$

The general solution of the differential equation $y dx + (1 + x^2) \tan^{-1} x dy = 0$ is

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