Solve the following initial value problems :
`x (dy)/(dx)+2y=x^(2)(x !=0)`, given that `y=0` when `x=1`.

Answer

To solve the initial value problem given by the differential equation

\[ x \frac{dy}{dx} + 2y = x^2 \quad (x \neq 0) \]

with the initial condition \(y(1) = 0\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
We start by dividing the entire equation by \(x\):

\[ \frac{dy}{dx} + \frac{2}{x}y = x \]

### Step 2: Identify \(p(x)\) and \(q(x)\)
From the standard form \(\frac{dy}{dx} + p(x)y = q(x)\), we identify:
- \(p(x) = \frac{2}{x}\)
- \(q(x) = x\)

### Step 3: Find the integrating factor
The integrating factor \(I(x)\) is given by:

\[ I(x) = e^{\int p(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = |x|^2 \]

Since \(x \neq 0\), we can simplify this to:

\[ I(x) = x^2 \]

### Step 4: Multiply the differential equation by the integrating factor
Now, we multiply the entire equation by the integrating factor \(x^2\):

\[ x^2 \frac{dy}{dx} + 2xy = x^3 \]

### Step 5: Recognize the left-hand side as a derivative
The left-hand side of the equation can be expressed as the derivative of a product:

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

### Step 6: Integrate both sides
Now we integrate both sides with respect to \(x\):

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

This gives us:

\[ x^2 y = \frac{x^4}{4} + C \]

### Step 7: Solve for \(y\)
Now, we can solve for \(y\):

\[ y = \frac{x^4}{4x^2} + \frac{C}{x^2} = \frac{x^2}{4} + \frac{C}{x^2} \]

### Step 8: Apply the initial condition
We use the initial condition \(y(1) = 0\):

\[ 0 = \frac{1^2}{4} + \frac{C}{1^2} \]

This simplifies to:

\[ 0 = \frac{1}{4} + C \implies C = -\frac{1}{4} \]

### Step 9: Substitute \(C\) back into the equation
Now we substitute \(C\) back into the equation for \(y\):

\[ y = \frac{x^2}{4} - \frac{1}{4x^2} \]

### Final Solution
Thus, the solution to the initial value problem is:

\[ y = \frac{x^2}{4} - \frac{1}{4x^2} \]