Let `y=f (x)` be a real valued function satisfying `x (dy)/(dx) =x ^(2) +y-2, f (1)=1,` then :

Options

A. `f (x)` is minimum at `x =1`
B. `f (x)` is maximum at `x=1`
C. `f (3)=5`
D. `f(2)=3`

Answer

To solve the differential equation given in the problem, we will follow these steps:

### Step 1: Write the given differential equation
The differential equation given is:
\[ x \frac{dy}{dx} = x^2 + y - 2 \]

### Step 2: Rearrange the equation
We can rearrange the equation 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 3: Rewrite in standard linear form
We can rewrite the equation in the standard linear form:
\[ \frac{dy}{dx} - \frac{y}{x} = x - \frac{2}{x} \]
Here, we identify \(p = -\frac{1}{x}\) and \(q = x - \frac{2}{x}\).

### Step 4: Find the integrating factor
The integrating factor \(I\) is given by:
\[ I = e^{\int p \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln |x|} = \frac{1}{x} \]

### Step 5: Multiply through by the integrating factor
Now, we multiply the entire equation by the integrating factor:
\[ \frac{1}{x} \frac{dy}{dx} - \frac{y}{x^2} = 1 - \frac{2}{x^2} \]

### Step 6: Rewrite the left-hand side
The left-hand side can be rewritten as the derivative of a product:
\[ \frac{d}{dx}\left(\frac{y}{x}\right) = 1 - \frac{2}{x^2} \]

### Step 7: Integrate both sides
Now we integrate both sides:
\[ \int \frac{d}{dx}\left(\frac{y}{x}\right) \, dx = \int \left(1 - \frac{2}{x^2}\right) \, dx \]
This gives:
\[ \frac{y}{x} = x + \frac{2}{x} + C \]
where \(C\) is the constant of integration.

### Step 8: Solve for \(y\)
Multiplying through by \(x\) to solve for \(y\):
\[ y = x^2 + 2 + Cx \]

### Step 9: Apply the initial condition
We are given that \(f(1) = 1\). Therefore:
\[ 1 = 1^2 + 2 + C(1) \]
This simplifies to:
\[ 1 = 1 + 2 + C \implies C = -2 \]

### Step 10: Write the particular solution
Substituting \(C\) back into the equation for \(y\):
\[ y = x^2 + 2 - 2x \]
Thus, the particular solution is:
\[ y = x^2 - 2x + 2 \]

### Step 11: Analyze the function for extrema
To find whether this function has a minimum or maximum at \(x = 1\), we compute the first derivative:
\[ \frac{dy}{dx} = 2x - 2 \]
Setting this equal to zero to find critical points:
\[ 2x - 2 = 0 \implies x = 1 \]

### Step 12: Compute the second derivative
Now we compute the second derivative:
\[ \frac{d^2y}{dx^2} = 2 \]
Since the second derivative is positive, this indicates that the function has a minimum at \(x = 1\).

### Step 13: Check the values at specific points
Now we check the values of \(f(3)\) and \(f(2)\):
1. For \(f(3)\):
\[ f(3) = 3^2 - 2(3) + 2 = 9 - 6 + 2 = 5 \]
2. For \(f(2)\):
\[ f(2) = 2^2 - 2(2) + 2 = 4 - 4 + 2 = 2 \]

### Conclusion
The final results are:
- \(f(3) = 5\) (Correct)
- \(f(2) = 2\) (Incorrect)

Thus, the correct options are:
- \(f(x)\) is minimum at \(x = 1\)
- \(f(3) = 5\)