Solve the following initial value problems :
`x(dy)/(dx)+2y=x^(2),y(1)=1/4`

Answer

To solve the initial value problem given by the differential equation \( x \frac{dy}{dx} + 2y = x^2 \) with the initial condition \( y(1) = \frac{1}{4} \), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ x \frac{dy}{dx} + 2y = x^2 \]
We can divide through by \( x \) (assuming \( x \neq 0 \)):
\[ \frac{dy}{dx} + \frac{2y}{x} = x \]

### Step 2: Identify \( p(x) \) and \( q(x) \)
This is now in the standard linear form:
\[ \frac{dy}{dx} + p(x)y = q(x) \]
where \( p(x) = \frac{2}{x} \) and \( q(x) = x \).

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int p(x) \, dx} = e^{\int \frac{2}{x} \, dx} = e^{2 \ln |x|} = |x|^2 \]
Since we are considering \( x > 0 \), we have:
\[ \mu(x) = x^2 \]

### Step 4: Multiply through by the integrating factor
We multiply the entire differential equation by the integrating factor:
\[ x^2 \frac{dy}{dx} + 2xy = x^3 \]

### Step 5: Rewrite the left-hand side as a derivative
The left-hand side can be rewritten as:
\[ \frac{d}{dx}(x^2 y) = x^3 \]

### Step 6: Integrate both sides
Integrating 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
Using the initial condition \( y(1) = \frac{1}{4} \):
\[ \frac{1}{4} = \frac{1^2}{4} + \frac{C}{1^2} \]
This simplifies to:
\[ \frac{1}{4} = \frac{1}{4} + C \]
Thus, \( C = 0 \).

### Step 9: Write the final solution
Substituting \( C \) back into the equation for \( y \):
\[ y = \frac{x^2}{4} \]

### Final Answer
The solution to the initial value problem is:
\[ y = \frac{x^2}{4} \]