Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2 ight) e^{ an x} d x+\cos ^2 x\left(1+e^{2 an x} ight) d y=0, y(0)=1$. Then $y\left(\frac{\pi}{4} ight)$ is equal to

Question

Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2ight) e^{	an x} d x+\cos ^2 x\left(1+e^{2 	an x}ight) d y=0, y(0)=1$. Then $y\left(\frac{\pi}{4}ight)$ is equal to

Aicharya · Accepted Answer

$\begin{aligned} & \left(1+y^2ight) e^{	an x} d x+\cos ^2 x\left(1+e^{2 	an x}ight) d y=0 \ & \int \frac{\sec ^2 x e^{	an x}}{1+e^{2 	an x}} d x+\int \frac{d y}{1+y^2}=C \ & \Rightarrow 	an ^{-1}\left(e^{	an x}ight)+	an ^{-1} y=C \ & 	ext { for } x=0, y=1, 	an ^{-1}(1)+	an ^{-1} 1=C\end{aligned}$ $\mathrm{C}=\frac{\pi}{2}$ $	an ^{-1}\left(e^{	an x}ight)+	an ^{-1} y=\frac{\pi}{2}$ Put $x=\pi, 	an ^{-1} e+	an ^{-1} y=\frac{\pi}{2}$ $	an ^{-1} y=\cot ^{-1} e$ $y=\frac{1}{e}$

Let $y=y(x)$ be the solution of the differential equation $\left(1+y^2\right) e^{\tan x} d x+\cos ^2…

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