Question

Solve ${\cos }^{2}x\frac{dy}{dx}+y=\tan x\left(0\le x<\frac{\pi }{2}\right)$

Answer 1

We have,

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

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

We know that the general equation

$\frac{dy}{dx}+Py=Q$

Here,

$P={\sec }^{2}x,Q={\sec }^{2}x\tan x$

Since, integrating factor

$I.F=e^{\int Pdx}$

$I.F=e^{\int {\sec }^{2}xdx}$

$I.F=e^{\tan x}$

We know that the general solution,

$y\times I.F=\int (Q\times I.F)dx+C$

Therefore,

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

Let $t=\tan x$

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

Therefore,

$y\times e^t=\int t{e}^{t}dt+C$

$y\times e^t=t{e}^t-\int 1e^{t}dt+C$

$y\times e^t=t{e}^t-e^t+C$

On putting the value of $t$, we get

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

Hence, this is the answer.