The general solution of the differential equa | vedclass

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(D) Given the differential equation: $(1+y^2) dx = (	an^{-1} y - x) dy$ Rearranging the terms, we get: $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}

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

Vedclass · Answer

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

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

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