Question

Solve the differential equation ${\cos }^{2}x\frac{dy}{dx}+y=\tan x$

Answer 1

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

$\frac{dy}{dx}+\frac{y}{{\cos }^{2}x}=\frac{\tan x}{{\cos }^{2}x}$ (1)

This is of the form :$\frac{dy}{dx}+P\left(x\right)y=Q\left(x\right)$

where $P\left(x\right)=\frac{1}{{\cos }^{2}x}={\sec }^{2}x$

$Q\left(x\right)=\tan x{\sec }^{2}x$

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

Mutiplying (1) by Integrating Factor:

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

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

Integerating both sides,

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

Taking RHS

Let $\tan x=t$

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

$\int \tan x{\sec }^2{e}^{\tan x}dx=\int t{e}^{t}dt=t\int e^t+\int e^t=(t+1)e^t$

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

Hence

$y\tan x=(\tan x+1)e^{\tan x}+c$ where c is constant of integration

$y=\frac{(\tan x+1)e^{\tan x}}{\tan x}+c\tan x$