Solution of the differential equation `cos x dy= y (sin x-y) dx, 0 lt x lt (pi)/(2)` is

Options

A. `sec x = (tan x +c)y`
B. `y sec x=tan x +c`
C. `y tan x=sec x +c`
D. `tan x=(sec x +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 equation
We start with the given equation:
\[ \cos x \, dy = y (\sin x - y) \, dx \]
We can rearrange this to:
\[ \frac{dy}{dx} = \frac{y (\sin x - y)}{\cos x} \]

### Step 2: Separate variables
Now, we can separate the variables \( y \) and \( x \):
\[ \frac{dy}{y (\sin x - y)} = \frac{dx}{\cos x} \]

### Step 3: Integrate both sides
Next, we integrate both sides. The left side requires partial fraction decomposition:
\[ \frac{1}{y (\sin x - y)} = \frac{A}{y} + \frac{B}{\sin x - y} \]
Multiplying through by the denominator \( y (\sin x - y) \) gives:
\[ 1 = A (\sin x - y) + B y \]
Setting \( y = 0 \) gives \( A \sin x = 1 \) so \( A = \frac{1}{\sin x} \). Setting \( y = \sin x \) gives \( B \sin x = 1 \) so \( B = \frac{1}{\sin x} \). Thus:
\[ \frac{1}{y (\sin x - y)} = \frac{1}{\sin x} \left( \frac{1}{y} + \frac{1}{\sin x - y} \right) \]
Integrating both sides:
\[ \int \left( \frac{1}{y} + \frac{1}{\sin x - y} \right) dy = \int \frac{dx}{\cos x} \]
The left side integrates to:
\[ \ln |y| - \ln |\sin x - y| = \ln \left| \frac{y}{\sin x - y} \right| \]
The right side integrates to:
\[ \ln |\sec x + \tan x| + C \]

### Step 4: Combine results
Combining the results gives:
\[ \ln \left| \frac{y}{\sin x - y} \right| = \ln |\sec x + \tan x| + C \]
Exponentiating both sides:
\[ \frac{y}{\sin x - y} = k (\sec x + \tan x) \]
where \( k = e^C \).

### Step 5: Solve for \( y \)
Now, we can solve for \( y \):
\[ y = k (\sec x + \tan x)(\sin x - y) \]
Rearranging gives:
\[ y + k (\sec x + \tan x) y = k (\sec x + \tan x) \sin x \]
Factoring out \( y \):
\[ y(1 + k (\sec x + \tan x)) = k (\sec x + \tan x) \sin x \]
Thus:
\[ y = \frac{k (\sec x + \tan x) \sin x}{1 + k (\sec x + \tan x)} \]

### Final Solution
The solution to the differential equation is:
\[ y = \frac{k (\sec x + \tan x) \sin x}{1 + k (\sec x + \tan x)} \]