Let y=y(x) be the solution of the differential equation [(1+y^2) e^ x d x+ ^2 x(1+e^2 x) d…

Question

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

Accepted Answer

$\begin{gathered} (1+{y}^{2}){e}^{	an x}dx+{\cos }^{2}x(1+{e}^{2	an x})dy=0 \ \frac{dy}{1+{y}^{2}}=-\frac{{e}^{	an x}\cdot {\sec }^{2}xdx}{1+{e}^{2	an x}} \ \int \frac{dy}{1+{y}^{2}}=-\int \frac{{e}^{	an x}\cdot {\sec }^{2}xdx}{1+{e}^{2	an x}} \end{gathered}$

Let ${e}^{	an x}=t$

$\begin{gathered} {e}^{	an x}\cdot {\sec }^{2}xdx=dt \ \int \frac{dy}{1+{y}^{2}}=-\int \frac{dt}{1+{t}^{2}} \end{gathered}$

$\begin{gathered} {	an }^{-1}y=-{	an }^{-1}t+c \ {	an }^{-1}y=-{	an }^{-1}({e}^{	an x})+c \ x=0,y=1 \ {	an }^{-1}(1)=-{	an }^{-1}(1)+c \ \frac{\pi }{4}=-\frac{\pi }{4}+c \ c=\frac{\pi }{2} \ {	an }^{-1}y=-{	an }^{-1}({e}^{	an x})+\frac{\pi }{2} \end{gathered}$

Now, at $x=\frac{\pi }{4}$

$\begin{gathered} {	an }^{-1}y=-{	an }^{-1}(e)+\frac{\pi }{2} \ {	an }^{-1}y={\cot }^{-1}e={	an }^{-1}\frac{1}{e} \ \Rightarrow y=\frac{1}{e} \end{gathered}$

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