Solution of the differential equation
`cosxdy=y(sinx-y)dx, 0ltxlt(pi)/(2)`is

Options

A. secx=(tanx+C)y
B. ysecx=tanx+C
C. ytanx=secx+C
D. tanx=(secx+C)y

Answer

To solve the differential equation \( \cos x \, dy = y(\sin x - y) \, dx \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
We start with the given equation:
\[ \cos x \, dy = y(\sin x - y) \, dx \]
Dividing both sides by \( dx \) gives:
\[ \cos x \frac{dy}{dx} = y(\sin x - y) \]

### Step 2: Rearranging the Equation
Rearranging the equation, we have:
\[ \frac{dy}{dx} = \frac{y(\sin x - y)}{\cos x} \]
This can be rewritten as:
\[ \frac{dy}{dx} = y \frac{\sin x}{\cos x} - \frac{y^2}{\cos x} \]
or
\[ \frac{dy}{dx} = y \tan x - y^2 \sec x \]

### Step 3: Separate Variables
Next, we separate the variables by dividing both sides by \( y^2 \):
\[ \frac{1}{y^2} \frac{dy}{dx} = \tan x - \sec x \cdot y \]
Let \( t = \frac{1}{y} \), then \( \frac{dy}{dx} = -\frac{1}{t^2} \frac{dt}{dx} \). Substituting this into the equation gives:
\[ -\frac{1}{t^2} \frac{dt}{dx} = \tan x - \sec x \cdot \frac{1}{t} \]
Multiplying through by \(-t^2\) results in:
\[ \frac{dt}{dx} = t \tan x - \sec x \]

### Step 4: Linear Differential Equation
This is a linear first-order differential equation in \( t \):
\[ \frac{dt}{dx} + t \tan x = \sec x \]
where \( p = \tan x \) and \( q = \sec x \).

### Step 5: Find the Integrating Factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int \tan x \, dx} = e^{-\ln(\cos x)} = \sec x \]

### Step 6: Solve the Differential Equation
Multiplying through by the integrating factor:
\[ \sec x \frac{dt}{dx} + t \sec x \tan x = \sec^2 x \]
The left side can be written as:
\[ \frac{d}{dx}(t \sec x) = \sec^2 x \]
Integrating both sides:
\[ t \sec x = \tan x + C \]
Thus,
\[ t = \frac{\tan x + C}{\sec x} \]
Substituting back \( t = \frac{1}{y} \):
\[ \frac{1}{y} \sec x = \tan x + C \]
or
\[ \sec x = y(\tan x + C) \]

### Step 7: Final Solution
Rearranging gives us the final solution:
\[ \sec x = y(\tan x + C) \]