$$ \cos ^ { 2 } x \frac { d y } { d x } + y = an x $$

Question

$$ \cos ^ { 2 } x \frac { d y } { d x } + y = 	an x $$

UpStudy AI · Accepted Answer

To solve the differential equation:

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

we can follow these steps:

### 1. Rewrite the Equation

First, let's rewrite the equation in the standard linear form:

$$
\frac{dy}{dx} + P(x) y = Q(x)
$$

Divide both sides by $$\cos^2 x$$:

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

Here,
$$
P(x) = \frac{1}{\cos^2 x} = \sec^2 x \quad \text{and} \quad Q(x) = \frac{\tan x}{\cos^2 x} = \frac{\sin x}{\cos^3 x}
$$

### 2. Find the Integrating Factor

The integrating factor $$\mu(x)$$ is given by:

$$
\mu(x) = e^{\int P(x) \, dx} = e^{\int \sec^2 x \, dx} = e^{\tan x}
$$

### 3. Multiply Through by the Integrating Factor

Multiply both sides of the differential equation by $$\mu(x)$$:

$$
e^{\tan x} \frac{dy}{dx} + e^{\tan x} \sec^2 x y = e^{\tan x} \frac{\sin x}{\cos^3 x}
$$

This simplifies to:

$$
\frac{d}{dx}\left( y e^{\tan x} \right) = e^{\tan x} \frac{\sin x}{\cos^3 x}
$$

### 4. Integrate Both Sides

Integrate both sides with respect to $$x$$:

$$
y e^{\tan x} = \int e^{\tan x} \frac{\sin x}{\cos^3 x} \, dx + C
$$

Let’s perform a substitution to evaluate the integral:

Let $$ u = \tan x $$, then $$ du = \sec^2 x \, dx $$.

Notice that:

$$
\frac{\sin x}{\cos^3 x} \, dx = \frac{\tan x}{\cos^2 x} \, dx = u \, du
$$

So the integral becomes:

$$
\int u e^{u} \, du = e^{u} (u - 1) + C = e^{\tan x} (\tan x - 1) + C
$$

### 5. Solve for $$ y $$

Substitute back to solve for $$ y $$:

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

Divide both sides by $$ e^{\tan x} $$:

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

### Final Solution

$$
\boxed{y = \tan x - 1 + C\, e^{ -\,\tan x}}
$$
The solution to the differential equation is: $$ y = \tan x - 1 + C\, e^{-\tan x} $$

\( \cos ^ { 2 } x \frac { d y } { d x } + y = \tan x \)