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 to the form $\frac{dx}{dy} + P(y)x = Q(y)$:$\frac{dx}{dy}

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

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(D) Given the differential equation:$(1+y^2) dx = (	an^{-1} y - x) dy$Rearranging the terms to the form $\frac{dx}{dy} + P(y)x = Q(y)$:$\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}$$\frac{dx}{dy} + \frac{1}{1+y^2} x = \frac{	an^{-1} y}{1+y^2}$This is a linear differential equation in $x$ where $P(y) = \frac{1}{1+y^2}$ and $Q(y) = \frac{	an^{-1} y}{1+y^2}$.The Integrating Factor $(IF)$ is:$IF = e^{\int P(y) dy} = e^{\int \frac{1}{1+y^2} dy} = e^{	an^{-1} y}$The general solution is given by:$x \cdot IF = \int Q(y) \cdot IF dy + k$$x e^{	an^{-1} y} = \int \frac{	an^{-1} y}{1+y^2} e^{	an^{-1} y} dy + k$Let $t = 	an^{-1} y$,then $dt = \frac{1}{1+y^2} dy$:$x e^{	an^{-1} y} = \int t e^t dt + k$Using integration by parts ($\int u dv = uv - \int v du$ with $u=t, dv=e^t dt$):$x e^{	an^{-1} y} = t e^t - e^t + k$$x e^{	an^{-1} y} = e^{	an^{-1} y} (	an^{-1} y - 1) + k$Dividing by $e^{	an^{-1} y}$:$x = (	an^{-1} y - 1) + k e^{-	an^{-1} y}$

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

Similar Questions

Observe the following statements:
$I$. If $dy+2xy dx=2e^{-x^2} dx$,then $ye^{x^2}=2x+c$
$II$. If $ye^{x^2}-2x=c$,then $dx=\frac{dy}{2e^{-x^2}-2xy}$
Which of the following is a correct statement?

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The solution of the differential equation $x\frac{dy}{dx} + y = x^2 + 3x + 2$ is

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