If the curve satisfies the differential equation `x.(dy)/(dx)=x^(2)+y-2` and passes through (1, 1), then is also passes through the point

Options

A. (4, 4)
B. (3, 3)
C. (2, 2)
D. (0, 0)

Answer

To solve the given differential equation and find the point through which the curve passes, we will follow these steps:

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

### Step 2: Rearrange the equation
We can rearrange the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \frac{x^2 + y - 2}{x} \]

### Step 3: Rewrite in standard linear form
We can rewrite the equation in the standard form of a linear differential equation:
\[ \frac{dy}{dx} - \frac{1}{x}y = \frac{x^2 - 2}{x} \]
Here, \(P(x) = -\frac{1}{x}\) and \(Q(x) = \frac{x^2 - 2}{x}\).

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

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

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

### Step 7: Integrate both sides
Integrating both sides:
\[ \int \frac{d}{dx}\left(\frac{y}{x}\right) \, dx = \int \left(1 - \frac{2}{x^2}\right) \, dx \]
This gives:
\[ \frac{y}{x} = x + \frac{2}{x} + C \]
Multiplying through by \(x\):
\[ y = x^2 + 2 - Cx \]

### Step 8: Use the initial condition to find C
We know that the curve passes through the point (1, 1):
\[ 1 = 1^2 + 2 - C(1) \]
This simplifies to:
\[ 1 = 1 + 2 - C \implies C = 2 \]

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

### Step 10: Check which points lie on the curve
Now we need to check which of the given points also lies on the curve. We will evaluate the equation \(y = x^2 - 2x + 2\) for each point.

1. **For (4, 4)**:
\[ y = 4^2 - 2(4) + 2 = 16 - 8 + 2 = 10 \quad (\text{not } 4) \]

2. **For (3, 3)**:
\[ y = 3^2 - 2(3) + 2 = 9 - 6 + 2 = 5 \quad (\text{not } 3) \]

3. **For (2, 2)**:
\[ y = 2^2 - 2(2) + 2 = 4 - 4 + 2 = 2 \quad (\text{is } 2) \]

4. **For (0, 0)**:
\[ y = 0^2 - 2(0) + 2 = 0 + 2 = 2 \quad (\text{not } 0) \]

### Conclusion
The point that also lies on the curve is \((2, 2)\).