Solution of diff. eqn. ` "tan"y (dy)/(dx) +tanx = cosy cos^(3)x` is …………

Answer

To solve the differential equation \( \tan y \frac{dy}{dx} + \tan x = \cos y \cos^3 x \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \tan y \frac{dy}{dx} + \tan x = \cos y \cos^3 x \]
We can rearrange it to isolate the derivative term:
\[ \tan y \frac{dy}{dx} = \cos y \cos^3 x - \tan x \]

### Step 2: Divide through by \( \cos y \)
Next, we divide the entire equation by \( \cos y \):
\[ \frac{\tan y}{\cos y} \frac{dy}{dx} = \cos^2 x - \frac{\tan x}{\cos y} \]
This simplifies to:
\[ \sec y \frac{dy}{dx} = \cos^2 x - \sec y \tan x \]

### Step 3: Substitute \( v = \sec y \)
Let’s make the substitution \( v = \sec y \). Then, differentiating \( v \) with respect to \( x \) gives:
\[ \frac{dv}{dx} = \sec y \tan y \frac{dy}{dx} \]
Thus, we can rewrite our equation as:
\[ \frac{dv}{dx} = \cos^2 x - v \tan x \]

### Step 4: Rearranging into standard form
Now we can rearrange the equation into the standard linear form:
\[ \frac{dv}{dx} + v \tan x = \cos^2 x \]

### Step 5: Identify \( p(x) \) and \( q(x) \)
Here, we identify:
- \( p(x) = \tan x \)
- \( q(x) = \cos^2 x \)

### Step 6: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \tan x \, dx} = e^{-\ln \cos x} = \sec x \]

### Step 7: Multiply through by the integrating factor
Multiply the entire differential equation by \( \sec x \):
\[ \sec x \frac{dv}{dx} + v \sec x \tan x = \sec x \cos^2 x \]
This simplifies to:
\[ \sec x \frac{dv}{dx} + v \sec x \tan x = \sec x \cos^2 x \]

### Step 8: Solve the left-hand side
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(v \sec x) = \sec x \cos^2 x \]

### Step 9: Integrate both sides
Integrate both sides:
\[ \int \frac{d}{dx}(v \sec x) \, dx = \int \sec x \cos^2 x \, dx \]
This gives:
\[ v \sec x = \int \sec x \cos^2 x \, dx + C \]

### Step 10: Solve the integral on the right-hand side
To solve \( \int \sec x \cos^2 x \, dx \), we can use the identity \( \cos^2 x = 1 - \sin^2 x \):
\[ \int \sec x \cos^2 x \, dx = \int \cos^2 x \, dx = \frac{1}{2}(x + \sin(2x)) + C \]

### Step 11: Substitute back for \( v \)
Substituting back for \( v \):
\[ \sec y \sec x = \frac{1}{2}(x + \sin(2x)) + C \]

### Step 12: Final solution
Thus, the final solution for the differential equation is:
\[ \sec y \sec x = \frac{1}{2}(x + \sin(2x)) + C \]