The equation of the curve passing through the point 0 , π 4 and satisfying the differential equation e…

$\Rightarrow e^x \tan ydx+\left(1+e^x\right)\left({\sec }^{2}y\right)dy=0$

$\Rightarrow e^x \tan ydx=-\left(1+e^x\right)\left({\sec }^{2}y\right)dy$

$\Rightarrow \frac{e^x}{\left(1+e^x\right)}dx=-\frac{{\sec }^{2}y }{\tan y}dy$

Integrate both sides

$\int \frac{e^x}{\left(1+e^x\right)}dx=-\int \frac{{\sec }^{2}y}{\tan y}dy$

Formula:$\int \frac{f^{\text{'}}\left(x\right)}{f\left(x\right)}dx=\log f\left(x\right)+c$

$\frac{d}{dy}\tan y={\sec }^{2}y$ & $\frac{d}{dx}\left(1+e^x\right)=e^x$

$\Rightarrow \log \left(1+e^x\right)=-\log \left(\tan y\right)+\log C$

$\Rightarrow \log \left(1+e^x\right)+\log \left(\tan y\right)=\log C$

$\Rightarrow \log [\left(1+e^x\right) \tan y]=\log C$

$\Rightarrow \left(1+e^x\right) \tan y=C$

The curve passing through the point$\left(x,y\right)=\left(0,\frac{\mathrm{\pi }}{4}\right)$

$\Rightarrow \left(1+e^0\right) \tan \left(\frac{\mathrm{\pi }}{4}\right)=C$

$\Rightarrow C=2$

$\Rightarrow \left(1+e^x\right) \tan y=2$