Question

Find the particular solution of the differential equation $e^x\tan ydx+(2-e^x){\sec }^{2}ydy=0,$ given that $y=\frac{\pi }{4}$ when $x=0$.

Answer 1

The given equation is
$e^x\tan ydx+(2-e^x){\sec }^{2}ydy=0$

$\Rightarrow -\frac{e^x}{2-e^x}dx=\frac{{\sec }^{2}y}{\tan y}dy$

$\Rightarrow -\int \frac{e^x}{2-e^x}dx=\int \frac{{\sec }^{2}y}{\tan y}dy$

Put $2-e^x=z\Rightarrow -e^{x}dx=dz$

And $\tan y=u\Rightarrow$ ${\sec }^{2}ydy=du$
$\therefore \int \frac{1}{z}dz=\int \frac{1}{u}du$

$\log z=\log u+\log C$ [$C$ is arbitary constant of intregation]

$\log z=\log (u\cdot C)$

$z=u\cdot C$

$\therefore 2-e^x=C\cdot \tan y$ ....... $\left(i\right)$

When $x=0$, then $y=\frac{\pi }{4}$ [given]

$\Rightarrow 2-e^0=C\cdot \tan \frac{\pi }{4}$

$\Rightarrow 2-1=C\cdot 1\Rightarrow C=1$

putting $C=1$ in the equation $\left(i\right)$

The particular solution of the given differential equation is,

$\tan y=2-e^x$