The solution of `(dy)/(dx)+y=e^(x)` is

Options

A. `2y=e^(2x)+c`
B. `2ye^(x)=e^(x)+c`
C. `2ye^(x)=e^(2x)+c`
D. `2ye^(2x)=2e^(x)+c`

Answer

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

### Step 1: Identify the form of the equation
The given equation is in the form of a linear first-order differential equation, which can be expressed as:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = 1\) and \(Q(x) = e^x\).

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

### Step 3: Multiply the entire equation by the integrating factor
Now we multiply the entire differential equation by the integrating factor \(e^x\):
\[ e^x \frac{dy}{dx} + e^x y = e^{2x} \]

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

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

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

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