Find the general solution of the following differential equations :
`y'+2y=e^(2x) `

Answer

To solve the differential equation \( y' + 2y = e^{2x} \), we will follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the equation in standard form:
\[ \frac{dy}{dx} + 2y = e^{2x} \]
Here, we identify \( p = 2 \) and \( q = e^{2x} \).

### 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
We multiply the entire differential equation by the integrating factor:
\[ e^{2x} \frac{dy}{dx} + 2e^{2x} y = e^{4x} \]

### Step 4: Rewrite the left-hand side as a derivative
The left-hand side can be expressed as the derivative of a product:
\[ \frac{d}{dx}(e^{2x} y) = e^{4x} \]

### Step 5: Integrate both sides
Now, we integrate both sides with respect to \( x \):
\[ \int \frac{d}{dx}(e^{2x} y) \, dx = \int e^{4x} \, dx \]
This gives us:
\[ e^{2x} y = \frac{1}{4} e^{4x} + C \]
where \( C \) is the constant of integration.

### Step 6: Solve for \( y \)
Now, we solve for \( y \):
\[ y = \frac{1}{4} e^{2x} + Ce^{-2x} \]

### Final Solution
Thus, the general solution of the differential equation is:
\[ y = \frac{1}{4} e^{2x} + Ce^{-2x} \]

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