The general solution of the differential equa | vedclass

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(A) Given differential equation is $(1+y^2) dx = (	an^{-1} y - x) dy$. Dividing by $(1+y^2) dy$, we get $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2

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$x = (	an^{-1} y) - 1 + C e^{-	an^{-1} y}$

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(A) Given differential equation is $(1+y^2) dx = (	an^{-1} y - x) dy$. Dividing by $(1+y^2) dy$, we get $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}$. Rearranging the terms, we have $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{	an^{-1} y}{1+y^2}$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$, where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{	an^{-1} y}{1+y^2}$. The integrating factor $IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an^{-1} y}$. The general solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + C$. $x \cdot e^{	an^{-1} y} = \int \frac{	an^{-1} y}{1+y^2} e^{	an^{-1} y} dy + C$. Let $t = 	an^{-1} y$, then $dt = \frac{1}{1+y^2} dy$. $x \cdot e^{	an^{-1} y} = \int t e^t dt + C$. Using integration by parts, $\int t e^t dt = t e^t - e^t$. $x \cdot e^{	an^{-1} y} = e^t(t - 1) + C$. Substituting $t = 	an^{-1} y$, we get $x \cdot e^{	an^{-1} y} = e^{	an^{-1} y}(	an^{-1} y - 1) + C$. Dividing by $e^{	an^{-1} y}$, we get $x = 	an^{-1} y - 1 + C e^{-	an^{-1} y}$.

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

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