Statement I Integral curves denoted by the first order linear differential equation `(dy)/(dx)-(1)/(x)y=-x` are family of parabolas passing throught the origin.
Statement II Every differential equation geomrtrically represents a family of curve having some common property.

Options

A. Statement I is true ,and Statement II is the correct explanation for Statement I.
B. Statement I is true, Statement II is true and Statement II is the correct explanation for Statment I
C. Statement I is true, Statement II is false.
D. Statement I is false, Statement II is true.

Answer

To solve the problem, we need to analyze the two statements provided regarding the first-order linear differential equation:

1. **Statement I**: The integral curves denoted by the first-order linear differential equation \(\frac{dy}{dx} - \frac{1}{x}y = -x\) are a family of parabolas passing through the origin.
2. **Statement II**: Every differential equation geometrically represents a family of curves having some common property.

### Step-by-Step Solution

**Step 1: Identify the differential equation.**
The given differential equation is:
\[ \frac{dy}{dx} - \frac{1}{x}y = -x \]

**Step 2: Determine if it is a linear differential equation.**
This is a first-order linear differential equation of the form:
\[ \frac{dy}{dx} + P(x)y = Q(x) \]
where \(P(x) = -\frac{1}{x}\) and \(Q(x) = -x\).

**Step 3: Find the integrating factor.**
The integrating factor \(I(x)\) is given by:
\[ I(x) = e^{\int P(x) \, dx} = e^{\int -\frac{1}{x} \, dx} = e^{-\ln|x|} = \frac{1}{|x|} \]
For simplicity, we can use \(I(x) = \frac{1}{x}\) (assuming \(x > 0\)).

**Step 4: Multiply the entire differential equation by the integrating factor.**
Multiplying through by \(\frac{1}{x}\):
\[ \frac{1}{x}\frac{dy}{dx} - \frac{1}{x^2}y = -1 \]

**Step 5: Rewrite the left-hand side as a derivative.**
The left-hand side can be rewritten as:
\[ \frac{d}{dx}\left(\frac{y}{x}\right) = -1 \]

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

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

**Step 8: Analyze the solution.**
The general solution \(y = -x^2 + Cx\) represents a family of parabolas. To check if they pass through the origin, substitute \(x = 0\):
\[ y(0) = -0^2 + C(0) = 0 \]
Thus, all curves pass through the origin.

### Conclusion
- **Statement I** is true: The integral curves are indeed parabolas passing through the origin.
- **Statement II** is also true: Every differential equation represents a family of curves with some common property.

### Final Answer
Both statements are true, and Statement II serves as a correct explanation for Statement I.