` (dy)/(dx) + 2xy = x`, given that `y =1` when ` x =0`

Explanation

To solve the differential equation \(\frac{dy}{dx} + 2xy = x\) with the initial condition \(y(0) = 1\), we will follow these steps:

### Step 1: Identify the standard form
The given equation is already in the standard form of a first-order linear differential equation:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = 2x\) and \(Q(x) = x\).

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

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

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

### Step 5: Integrate both sides
Now we integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx}(e^{x^2} y) \, dx = \int e^{x^2} x \, dx \]
The left-hand side simplifies to:
\[ e^{x^2} y \]
For the right-hand side, we can use integration by parts or recognize that:
\[ \int e^{x^2} x \, dx = \frac{1}{2} e^{x^2} + C \]
Thus, we have:
\[ e^{x^2} y = \frac{1}{2} e^{x^2} + C \]

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

### Step 7: Apply the initial condition
We use the initial condition \(y(0) = 1\):
\[ 1 = \frac{1}{2} + Ce^{0} \]
This simplifies to:
\[ 1 = \frac{1}{2} + C \implies C = \frac{1}{2} \]

### Step 8: Write the final solution
Substituting \(C\) back into the equation for \(y\):
\[ y = \frac{1}{2} + \frac{1}{2} e^{-x^2} \]

### Final Answer:
\[ y = \frac{1}{2} (1 + e^{-x^2}) \]

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