Solve the differential equation: `(dy)/(dx)-y=cos x`

Explanation

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

### Step 1: Identify the form of the differential equation
The given equation is in the standard linear form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = -1\) and \(Q(x) = \cos x\).

### Step 2: Find the integrating factor
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int P(x) \, dx} = e^{\int -1 \, dx} = e^{-x} \]

### Step 3: Multiply the entire differential equation by the integrating factor
Multiplying the original equation by \(e^{-x}\):
\[ e^{-x} \frac{dy}{dx} - e^{-x} y = e^{-x} \cos x \]

### Step 4: Rewrite the left-hand side as a derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(e^{-x} y) = e^{-x} \cos x \]

### Step 5: Integrate both sides
Now, we integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx}(e^{-x} y) \, dx = \int e^{-x} \cos x \, dx \]
The left-hand side simplifies to:
\[ e^{-x} y = \int e^{-x} \cos x \, dx \]

### Step 6: Solve the integral on the right-hand side
To solve the integral \(\int e^{-x} \cos x \, dx\), we use integration by parts. Let:
- \(u = \cos x\) and \(dv = e^{-x} dx\)
Then, \(du = -\sin x \, dx\) and \(v = -e^{-x}\).

Using integration by parts:
\[ \int e^{-x} \cos x \, dx = -e^{-x} \cos x - \int -e^{-x} (-\sin x) \, dx \]
This simplifies to:
\[ -e^{-x} \cos x + \int e^{-x} \sin x \, dx \]

Now we need to solve \(\int e^{-x} \sin x \, dx\) using integration by parts again:
Let:
- \(u = \sin x\) and \(dv = e^{-x} dx\)
Then, \(du = \cos x \, dx\) and \(v = -e^{-x}\).

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

### Step 7: Combine the results
Let \(I = \int e^{-x} \cos x \, dx\). We have:
\[ I = -e^{-x} \cos x + (-e^{-x} \sin x + I) \]
This leads to:
\[ 2I = -e^{-x} (\cos x + \sin x) \]
Thus:
\[ I = -\frac{1}{2} e^{-x} (\cos x + \sin x) \]

### Step 8: Substitute back into the equation
Substituting \(I\) back, we have:
\[ e^{-x} y = -\frac{1}{2} e^{-x} (\cos x + \sin x) + C \]
Multiplying through by \(e^x\):
\[ y = -\frac{1}{2} (\cos x + \sin x) + Ce^x \]

### Final Solution
The solution to the differential equation is:
\[ y = -\frac{1}{2} (\cos x + \sin x) + Ce^x \]