Question

solve the differential equation : $\frac{dy}{dx}+y{\sec }^{2}x=\tan x{\sec }^{2}y;y\left(0\right)=1$

Answer 1

$\frac{dy}{dx}+y{\sec }^{2}x=\tan x.{\sec }^{2}y$ $y\left(0\right)=1$

Integrating Factor,

$e^{\int {\sec }^{2}xdx}=e^{\tan x}$

solution,

$y(I.F)=\int \tan x.se{c}^{2}x(I.F)dx$ ......(1)

$y.e^{\tan x}=\int \tan x.{\sec }^{2}x.e^{\tan x}.dx$

$\int \tan x.{\sec }^{2}x.e^{\tan x}dx$

Let $\tan x=t$

$\Rightarrow {\sec }^{2}xdx=dt$

$\int \tan x.{\sec }^{2}x.e^{\tan x}dx=\int t.e^{t}dt$

$=t{e}^t-e^t$

$=(\tan x-1)e^{tanx}$

$\therefore$ solution $\to y{e}^{\tan x}\Rightarrow (\tan x-1)e^{\tan x}+C$

Now when $x=0$ $y=1$

$1=-1+c\Rightarrow C=2$

$\therefore y{e}^{\tan x}=\tan x.e^{\tan x}-e^{\tan x}+2$