Solve the differential equation frac{dy}{dx} | vedclass

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Question

(B) Given the differential equation $\frac{dy}{dx} = \frac{1+y^2}{	an^{-1} y - x}$.Taking the reciprocal,we get $\frac{dx}{dy} = \frac{	an^{-1} y -

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) Given the differential equation $\frac{dy}{dx} = \frac{1+y^2}{	an^{-1} y - x}$.Taking the reciprocal,we get $\frac{dx}{dy} = \frac{	an^{-1} y - x}{1+y^2}$.Rearranging the terms,we have $\frac{dx}{dy} + \frac{x}{1+y^2} = \frac{	an^{-1} y}{1+y^2}$,which is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$.Here,$P(y) = \frac{1}{1+y^2}$ 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 solution is given by $x \cdot IF = \int Q(y) \cdot IF dy + c$.Substituting the values: $x e^{	an^{-1} y} = \int \frac{	an^{-1} y}{1+y^2} e^{	an^{-1} y} dy + c$.Let $t = 	an^{-1} y$,then $dt = \frac{1}{1+y^2} dy$.The integral becomes $\int t e^t dt + c$.Using integration by parts: $\int t e^t dt = t e^t - \int e^t dt = t e^t - e^t + c = e^t(t - 1) + c$.Substituting $t = 	an^{-1} y$ back,we get $x e^{	an^{-1} y} = e^{	an^{-1} y}(	an^{-1} y - 1) + c$.Thus,option $B$ is correct.

Solve the differential equation $\frac{dy}{dx} = \frac{1+y^2}{(\tan^{-1} y) - x}$.

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