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(B) Given differential equation is $(1+e^{2x}) dy + e^x(1+y^2) dx = 0$. Rearranging the terms to separate variables,we get $\frac{dy}{1+y^2} = -\frac{

Vedclass · Accepted Answer

$	an^{-1} e^x + 	an^{-1} y = \frac{\pi}{2}$

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(B) Given differential equation is $(1+e^{2x}) dy + e^x(1+y^2) dx = 0$. Rearranging the terms to separate variables,we get $\frac{dy}{1+y^2} = -\frac{e^x}{1+e^{2x}} dx$. Integrating both sides,$\int \frac{dy}{1+y^2} = -\int \frac{e^x}{1+(e^x)^2} dx$. Let $e^x = t$,then $e^x dx = dt$. The integral becomes $	an^{-1}(y) = -	an^{-1}(t) + C$,which is $	an^{-1}(y) = -	an^{-1}(e^x) + C$. Thus,$	an^{-1}(y) + 	an^{-1}(e^x) = C$. Given $x=0$ and $y=1$,we substitute these values: $	an^{-1}(1) + 	an^{-1}(e^0) = C$. $\frac{\pi}{4} + \frac{\pi}{4} = C$,so $C = \frac{\pi}{2}$. The particular solution is $	an^{-1}(y) + 	an^{-1}(e^x) = \frac{\pi}{2}$.

The particular solution of the differential equation $(1+e^{2x}) dy + e^x(1+y^2) dx = 0$ at $x=0$ and $y=1$ is

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