Solve the following differential equations
`(y- sin x)dx + tan x dy=0, y(0)=0`.

Answer

To solve the differential equation \((y - \sin x)dx + \tan x \, dy = 0\) with the initial condition \(y(0) = 0\), we will follow these steps:

### Step 1: Rearranging the Equation
We start with the given equation:
\[ (y - \sin x)dx + \tan x \, dy = 0 \]
We can rearrange it as:
\[ (y - \sin x)dx = -\tan x \, dy \]
Dividing both sides by \(\tan x\):
\[ \frac{dy}{dx} = \frac{y - \sin x}{-\tan x} = \frac{\sin x - y}{\tan x} \]

### Step 2: Simplifying the Equation
We know that \(\tan x = \frac{\sin x}{\cos x}\), so:
\[ \frac{dy}{dx} = \frac{\sin x - y}{\frac{\sin x}{\cos x}} = (\sin x - y) \cdot \frac{\cos x}{\sin x} \]
This simplifies to:
\[ \frac{dy}{dx} + y \cdot \cot x = \cos x \]
where \(\cot x = \frac{\cos x}{\sin x}\).

### Step 3: Identifying p and q
We can see that this is a linear first-order differential equation in the standard form:
\[ \frac{dy}{dx} + p(x)y = q(x) \]
where \(p(x) = \cot x\) and \(q(x) = \cos x\).

### Step 4: Finding the Integrating Factor
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int p(x) \, dx} = e^{\int \cot x \, dx} \]
The integral of \(\cot x\) is \(\ln |\sin x|\), so:
\[ I(x) = e^{\ln |\sin x|} = |\sin x| \]
Since \(\sin x\) is positive in the interval we are considering, we can write:
\[ I(x) = \sin x \]

### Step 5: Multiplying the Equation by the Integrating Factor
Now we multiply the entire differential equation by the integrating factor:
\[ \sin x \frac{dy}{dx} + y \sin x \cot x = \sin x \cos x \]
This simplifies to:
\[ \frac{d}{dx}(y \sin x) = \sin x \cos x \]

### Step 6: Integrating Both Sides
Now we integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx}(y \sin x) \, dx = \int \sin x \cos x \, dx \]
The left side gives:
\[ y \sin x = \int \sin x \cos x \, dx \]
Using the identity \(\sin x \cos x = \frac{1}{2} \sin(2x)\):
\[ \int \sin x \cos x \, dx = \frac{1}{2} \int \sin(2x) \, dx = -\frac{1}{4} \cos(2x) + C \]
Thus, we have:
\[ y \sin x = -\frac{1}{4} \cos(2x) + C \]

### Step 7: Solving for y
Now, we can solve for \(y\):
\[ y = \frac{-\frac{1}{4} \cos(2x) + C}{\sin x} \]

### Step 8: Applying the Initial Condition
We apply the initial condition \(y(0) = 0\):
\[ 0 = \frac{-\frac{1}{4} \cos(0) + C}{\sin(0)} \]
Since \(\sin(0) = 0\), we need to evaluate the limit as \(x\) approaches \(0\):
\[ \lim_{x \to 0} y = \lim_{x \to 0} \frac{-\frac{1}{4} \cos(2x) + C}{\sin x} \]
Using L'Hôpital's Rule:
\[ \lim_{x \to 0} \frac{-\frac{1}{4}(-2 \sin(2x))}{\cos x} = \lim_{x \to 0} \frac{\frac{1}{2} \sin(2x)}{\cos x} = 0 \]
This implies:
\[ C = \frac{1}{4} \]

### Final Solution
Substituting \(C\) back into our equation:
\[ y = \frac{-\frac{1}{4} \cos(2x) + \frac{1}{4}}{\sin x} = \frac{1 - \cos(2x)}{4 \sin x} \]
Using the identity \(1 - \cos(2x) = 2 \sin^2 x\):
\[ y = \frac{2 \sin^2 x}{4 \sin x} = \frac{\sin x}{2} \]

Thus, the solution to the differential equation is:
\[ \boxed{y = \frac{\sin x}{2}} \]