`(1+x)(dy)/(dx)=((1+x)^2+(y-3))`, If `y(2)=0` then `y(3)=?`

Explanation

To solve the differential equation given by

\[ (1+x) \frac{dy}{dx} = (1+x)^2 + (y-3) \]

with the initial condition \( y(2) = 0 \), we will follow these steps:

### Step 1: Rearranging the Equation
First, we can rearrange the equation to isolate \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = \frac{(1+x)^2 + (y-3)}{1+x} \]

### Step 2: Simplifying the Right Side
Now, we can simplify the right side:

\[ \frac{dy}{dx} = (1+x) + \frac{y-3}{1+x} \]

This gives us:

\[ \frac{dy}{dx} = 1 + x + \frac{y-3}{1+x} \]

### Step 3: Writing in Standard Form
Next, we can write this in the standard form of a linear first-order differential equation:

\[ \frac{dy}{dx} - \frac{1}{1+x} y = 1 + x - \frac{3}{1+x} \]

### Step 4: Identifying \(p\) and \(q\)
Here, we identify \(p\) and \(q\):

- \(p = -\frac{1}{1+x}\)
- \(q = 1 + x - \frac{3}{1+x}\)

### Step 5: Finding the Integrating Factor
The integrating factor \(I\) is given by:

\[ I = e^{\int p \, dx} = e^{\int -\frac{1}{1+x} \, dx} = e^{-\ln(1+x)} = \frac{1}{1+x} \]

### Step 6: Multiplying through by the Integrating Factor
Now, we multiply the entire differential equation by the integrating factor:

\[ \frac{1}{1+x} \frac{dy}{dx} - \frac{1}{(1+x)^2} y = \frac{1 + x - \frac{3}{1+x}}{1+x} \]

This simplifies to:

\[ \frac{d}{dx} \left( \frac{y}{1+x} \right) = \frac{1+x}{1+x} - \frac{3}{(1+x)^2} = 1 - \frac{3}{(1+x)^2} \]

### Step 7: Integrating Both Sides
Now we integrate both sides:

\[ \int \frac{d}{dx} \left( \frac{y}{1+x} \right) dx = \int \left( 1 - \frac{3}{(1+x)^2} \right) dx \]

The left side gives:

\[ \frac{y}{1+x} \]

The right side integrates to:

\[ x + \frac{3}{1+x} + C \]

### Step 8: Solving for \(y\)
Thus, we have:

\[ \frac{y}{1+x} = x + \frac{3}{1+x} + C \]

Multiplying through by \(1+x\):

\[ y = (1+x) \left( x + \frac{3}{1+x} + C \right) \]

This simplifies to:

\[ y = x(1+x) + 3 + C(1+x) \]

### Step 9: Applying Initial Condition
Now we apply the initial condition \(y(2) = 0\):

\[ 0 = 2(1+2) + 3 + C(1+2) \]

This gives:

\[ 0 = 6 + 3 + 3C \implies 3C = -9 \implies C = -3 \]

### Step 10: Final Equation for \(y\)
Substituting \(C\) back into the equation for \(y\):

\[ y = x(1+x) + 3 - 3(1+x) \]

This simplifies to:

\[ y = x^2 + x + 3 - 3 - 3x = x^2 - 2x \]

### Step 11: Finding \(y(3)\)
Now we find \(y(3)\):

\[ y(3) = 3^2 - 2 \cdot 3 = 9 - 6 = 3 \]

Thus, the final answer is:

\[ \boxed{3} \]