Solve the following differential equations :
`(dy)/(dx)=y-2 sin x`.

Answer

To solve the differential equation \(\frac{dy}{dx} = y - 2 \sin x\), we will follow the steps outlined below:

### Step 1: Rewrite the Equation
We start by rewriting the differential equation in standard form:
\[ \frac{dy}{dx} - y = -2 \sin x \]

### Step 2: Identify \(p\) and \(q\)
In the standard form \(\frac{dy}{dx} + p y = q\), we identify:
- \(p = -1\)
- \(q = -2 \sin x\)

### Step 3: Find the Integrating Factor
The integrating factor \(I\) is given by:
\[ I = e^{\int p \, dx} = e^{\int -1 \, dx} = e^{-x} \]

### Step 4: Multiply the Equation by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor \(e^{-x}\):
\[ e^{-x} \frac{dy}{dx} - e^{-x} y = -2 e^{-x} \sin x \]

### Step 5: Simplify the Left Side
The left-hand side can be simplified using the product rule:
\[ \frac{d}{dx}(y e^{-x}) = -2 e^{-x} \sin x \]

### Step 6: Integrate Both Sides
Next, we integrate both sides:
\[ \int \frac{d}{dx}(y e^{-x}) \, dx = \int -2 e^{-x} \sin x \, dx \]
This gives us:
\[ y e^{-x} = \int -2 e^{-x} \sin x \, dx + C \]

### Step 7: Solve the Integral
To solve the integral \(\int -2 e^{-x} \sin x \, dx\), we will use integration by parts. Let:
- \(u = \sin x\) and \(dv = -2 e^{-x} dx\)

Then, we differentiate and integrate:
- \(du = \cos x \, dx\)
- \(v = -2(-e^{-x}) = 2 e^{-x}\)

Using integration by parts:
\[ \int u \, dv = uv - \int v \, du \]
So we have:
\[ \int -2 e^{-x} \sin x \, dx = 2 e^{-x} \sin x - \int 2 e^{-x} \cos x \, dx \]

We will need to apply integration by parts again to solve \(\int 2 e^{-x} \cos x \, dx\).

### Step 8: Solve the Second Integral
Let:
- \(u = \cos x\) and \(dv = 2 e^{-x} dx\)

Then:
- \(du = -\sin x \, dx\)
- \(v = 2 e^{-x}\)

Thus:
\[ \int 2 e^{-x} \cos x \, dx = 2 e^{-x} \cos x - \int -2 e^{-x} \sin x \, dx \]

### Step 9: Combine the Results
Let \(I = \int -2 e^{-x} \sin x \, dx\). We have:
\[ I = 2 e^{-x} \sin x - (2 e^{-x} \cos x - I) \]
This simplifies to:
\[ 2I = 2 e^{-x} \sin x - 2 e^{-x} \cos x \]
Thus:
\[ I = e^{-x} (\sin x - \cos x) \]

### Step 10: Substitute Back
Substituting back into our equation gives:
\[ y e^{-x} = e^{-x} (\sin x - \cos x) + C \]
Multiplying through by \(e^{x}\):
\[ y = \sin x - \cos x + C e^{x} \]

### Final Solution
The final solution to the differential equation is:
\[ y = \sin x - \cos x + C e^{x} \]