If `x(dy)/(dx)=x^(2)+y-2, y(1)=1`, then `y(2)` equals ………………..

Explanation

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ x \frac{dy}{dx} = x^2 + y - 2 \]
Now, we can rearrange it to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{x^2 + y - 2}{x} \]
This simplifies to:
\[ \frac{dy}{dx} = x + \frac{y}{x} - \frac{2}{x} \]

### Step 2: Rearrange into standard linear form
Next, we rearrange the equation into the standard form of a linear differential equation:
\[ \frac{dy}{dx} - \frac{1}{x} y = x - \frac{2}{x} \]

### Step 3: Find the integrating factor
The integrating factor \( \mu(x) \) is given by:
\[ \mu(x) = e^{\int -\frac{1}{x} dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 4: Multiply through by the integrating factor
Now, we multiply the entire equation by the integrating factor \( \frac{1}{x} \):
\[ \frac{1}{x} \frac{dy}{dx} - \frac{1}{x^2} y = 1 - \frac{2}{x^2} \]

### Step 5: Rewrite the left side as a derivative
The left side can be rewritten as the derivative of a product:
\[ \frac{d}{dx} \left( \frac{y}{x} \right) = 1 - \frac{2}{x^2} \]

### Step 6: Integrate both sides
Integrating both sides with respect to \( x \):
\[ \int \frac{d}{dx} \left( \frac{y}{x} \right) dx = \int \left( 1 - \frac{2}{x^2} \right) dx \]
This gives us:
\[ \frac{y}{x} = x + \frac{2}{x} + C \]
where \( C \) is the constant of integration.

### Step 7: Solve for \( y \)
Multiplying through by \( x \):
\[ y = x^2 + 2 + Cx \]

### Step 8: Apply the initial condition
Now we use the initial condition \( y(1) = 1 \):
\[ 1 = 1^2 + 2 + C(1) \]
This simplifies to:
\[ 1 = 1 + 2 + C \implies C = 1 - 3 = -2 \]

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

### Step 10: Find \( y(2) \)
Now we need to find \( y(2) \):
\[ y(2) = 2^2 + 2 - 2(2) = 4 + 2 - 4 = 2 \]

### Final Answer
Thus, \( y(2) = 2 \).

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