Question

Solve: ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$.

Answer 1

Given equation is ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$

$\Rightarrow \frac{dy}{dx}+{\sec }^{2}xy=\tan x.se{c}^{2}x$

The genral solution of the equation,

$\frac{dy}{dx}+f\left(x\right).y=g\left(x\right)$

is $y=\left(e^{-\int f\left(x\right).dx}\right)(\int g(x)e^{\int f\left(x\right).dx}.dx+C)$

In the equation, $f\left(x\right)=se{c}^{2}x,g\left(x\right)=\tan x-{\sec }^{2}x$

The general equation is,

$y=\left(e^{-\int {\sec }^{2}xdx}\right)(\int \tan x.{\sec }^{2}x{e}^{\tan x}dx+C)$

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

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

$\therefore$ The solution is $y=e^{-\tan x}\left(e^{\tan x}\right(\tan x-1)+C)$

$y=\tan x-1+C{e}^{-\tan x}$