The general solution of the differential equa | vedclass

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

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$x \cdot e^{	an ^{-1} y}=\frac{(e^{	an ^{-1} y})^{2}}{2}+c$

Vedclass · Answer

(A) Given the differential equation: $(1+y^{2})+(x-e^{	an ^{-1} y}) \frac{dy}{dx}=0$ Rearranging the terms: $(x-e^{	an ^{-1} y}) \frac{dy}{dx} = -(1+y^{2})$ Taking the reciprocal: $\frac{dx}{dy} = \frac{-(x-e^{	an ^{-1} y})}{1+y^{2}} = \frac{-x}{1+y^{2}} + \frac{e^{	an ^{-1} y}}{1+y^{2}}$ 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 $Q(y) = \frac{e^{	an ^{-1} y}}{1+y^{2}}$. The Integrating Factor ($I$.$F$.) is $e^{\int P(y) dy} = e^{\int \frac{1}{1+y^{2}} dy} = e^{	an ^{-1} y}$. The general solution is given by $x \cdot (I.F.) = \int Q(y) \cdot (I.F.) dy + c$. $x \cdot e^{	an ^{-1} y} = \int \frac{e^{	an ^{-1} y}}{1+y^{2}} \cdot e^{	an ^{-1} y} dy + c$. Let $t = e^{	an ^{-1} y}$,then $dt = \frac{e^{	an ^{-1} y}}{1+y^{2}} dy$. Substituting this into the integral: $x \cdot e^{	an ^{-1} y} = \int t dt + c = \frac{t^{2}}{2} + c$. Thus,the general solution is $x \cdot e^{	an ^{-1} y} = \frac{(e^{	an ^{-1} y})^{2}}{2} + c$.

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

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