The solution of the differential equaiton
`cosxdy=y(sinx-y)dx` ,`0ltxltpi/2`

Options

A. `y tan x =sec x + C,` is
B. `tan x=(sec x+C)y`
C. `sec x=(tan x+C)y`
D. `y secx= tan x +C`

Answer

To solve the differential equation given by

\[ \cos x \, dy = y(\sin x - y) \, dx, \]

we will follow a systematic approach.

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate \( dy \) and \( dx \):

\[ \cos x \, dy = y(\sin x - y) \, dx. \]

Dividing both sides by \( y(\sin x - y) \):

\[ \frac{dy}{y(\sin x - y)} = \frac{\cos x}{\sin x} \, dx. \]

### Step 2: Separating Variables
Next, we can separate the variables:

\[ \frac{1}{y(\sin x - y)} \, dy = \frac{\cos x}{\sin x} \, dx. \]

### Step 3: Integrating Both Sides
Now, we will 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 \( y(\sin x - y) \) gives:

\[ 1 = A(\sin x - y) + By. \]

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, we can rewrite:

\[ \frac{1}{y(\sin x - y)} = \frac{1}{\sin x} \left( \frac{1}{y} + \frac{1}{\sin x - y} \right). \]

Now, integrating both sides:

\[ \int \left( \frac{1}{y} + \frac{1}{\sin x - y} \right) dy = \int \frac{\cos x}{\sin x} \, dx. \]

### Step 4: Solving the Integrals
The left side integrates to:

\[ \ln |y| - \ln |\sin x - y| = \ln \left| \frac{y}{\sin x - y} \right|. \]

The right side integrates to:

\[ \ln |\sin x| + C. \]

### Step 5: Equating Both Sides
Equating both sides gives:

\[ \ln \left| \frac{y}{\sin x - y} \right| = \ln |\sin x| + C. \]

Exponentiating both sides results in:

\[ \frac{y}{\sin x - y} = k \sin x, \]

where \( k = e^C \).

### Step 6: Solving for \( y \)
From the equation above, we can solve for \( y \):

\[ y = k \sin x (\sin x - y). \]

Rearranging gives:

\[ y + k y \sin x = k \sin^2 x. \]

Factoring out \( y \):

\[ y(1 + k \sin x) = k \sin^2 x. \]

Thus,

\[ y = \frac{k \sin^2 x}{1 + k \sin x}. \]

### Final Solution
The solution of the differential equation is:

\[ y = \frac{k \sin^2 x}{1 + k \sin x}. \]