Solve the following differential equations :
`(dy)/(dx)+2y=cos 3x`

Answer

To solve the differential equation

\[ \frac{dy}{dx} + 2y = \cos(3x), \]

we will follow these steps:

### Step 1: Identify p and q
We can rewrite the equation in the standard form:

\[ \frac{dy}{dx} + py = q, \]

where \( p = 2 \) and \( q = \cos(3x) \).

### Step 2: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:

\[ \mu(x) = e^{\int p \, dx} = e^{\int 2 \, dx} = e^{2x}. \]

### Step 3: Multiply the entire equation by the integrating factor
Now, we multiply the entire differential equation by the integrating factor \( e^{2x} \):

\[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{2x} \cos(3x). \]

### Step 4: Rewrite the left-hand side
The left-hand side can be rewritten as the derivative of a product:

\[ \frac{d}{dx}(e^{2x} y) = e^{2x} \cos(3x). \]

### Step 5: Integrate both sides
Now, we integrate both sides with respect to \( x \):

\[ \int \frac{d}{dx}(e^{2x} y) \, dx = \int e^{2x} \cos(3x) \, dx. \]

The left-hand side simplifies to:

\[ e^{2x} y = \int e^{2x} \cos(3x) \, dx. \]

### Step 6: Solve the integral on the right-hand side
To solve \( \int e^{2x} \cos(3x) \, dx \), we can use integration by parts or a known formula for integrals of the form \( e^{ax} \cos(bx) \):

\[ \int e^{ax} \cos(bx) \, dx = \frac{e^{ax}}{a^2 + b^2} (a \cos(bx) + b \sin(bx)). \]

Here, \( a = 2 \) and \( b = 3 \):

\[ \int e^{2x} \cos(3x) \, dx = \frac{e^{2x}}{2^2 + 3^2} (2 \cos(3x) + 3 \sin(3x)) = \frac{e^{2x}}{13} (2 \cos(3x) + 3 \sin(3x)). \]

### Step 7: Substitute back into the equation
Now substituting back, we have:

\[ e^{2x} y = \frac{e^{2x}}{13} (2 \cos(3x) + 3 \sin(3x)) + C, \]

where \( C \) is the constant of integration.

### Step 8: Solve for \( y \)
To find \( y \), we divide both sides by \( e^{2x} \):

\[ y = \frac{1}{13} (2 \cos(3x) + 3 \sin(3x)) + Ce^{-2x}. \]

### Final Solution
Thus, the solution to the differential equation is:

\[ y = \frac{2}{13} \cos(3x) + \frac{3}{13} \sin(3x) + Ce^{-2x}. \]