Let `y_(1) and y_(2)` be the two solutions of the differential equation `(dy)/(dx)+Py=Q`, where P and Q are functions of x,
Statement-1: The linear combination `ay_(1)+by_(2)` will be a solution of the differential equation, if `a+b=1`.
Statement-2 : The general solution of the differential equation is `y=y_(1)+C(y_(1)-y_(2))`.

Options

A. Statement-1 is True, Statement-2 is True, Statement-2 is a correct explanation for Statement-1.
B. Statement-1 is True, Statement-2 is True, Statement-2 is not a correct explanation for Statement-1.
C. Statement-1 is True, Statement-2 is False.
D. Statement-1 is False, Statement-2 is True.

Explanation

To solve the given problem, we need to analyze the statements about the solutions of the differential equation \(\frac{dy}{dx} + Py = Q\), where \(P\) and \(Q\) are functions of \(x\).

### Step-by-Step Solution:

1. **Understanding the Differential Equation**:
The given differential equation is:
\[ \frac{dy}{dx} + Py = Q \]
Here, \(y_1\) and \(y_2\) are two solutions of this equation.

2. **Verifying Statement 1**:
Statement 1 claims that the linear combination \(ay_1 + by_2\) will be a solution of the differential equation if \(a + b = 1\).

- Since \(y_1\) and \(y_2\) are solutions, we have:
\[ \frac{dy_1}{dx} + Py_1 = Q \]
\[ \frac{dy_2}{dx} + Py_2 = Q \]
- Let \(y = ay_1 + by_2\). Then, differentiating \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = a\frac{dy_1}{dx} + b\frac{dy_2}{dx} \]
- Substituting \(y\) into the differential equation:
\[ \frac{dy}{dx} + P(ay_1 + by_2) = a\frac{dy_1}{dx} + b\frac{dy_2}{dx} + P(ay_1 + by_2) \]
- Since \(P(ay_1 + by_2) = aPy_1 + bPy_2\), we can write:
\[ a\frac{dy_1}{dx} + b\frac{dy_2}{dx} + aPy_1 + bPy_2 = aQ + bQ = Q(a + b) \]
- If \(a + b = 1\), then:
\[ Q(a + b) = Q \]
- Thus, \(y = ay_1 + by_2\) satisfies the differential equation, confirming that Statement 1 is true.

3. **Verifying Statement 2**:
Statement 2 claims that the general solution of the differential equation is:
\[ y = y_1 + C(y_1 - y_2) \]
- We can express this as:
\[ y = y_1 + C(y_1 - y_2) = (1 + C)y_1 - Cy_2 \]
- Let \(a = 1 + C\) and \(b = -C\). Then:
\[ a + b = (1 + C) - C = 1 \]
- This shows that the general solution can also be expressed as a linear combination of \(y_1\) and \(y_2\) where \(a + b = 1\), confirming that Statement 2 is also true.

### Conclusion:
Both statements are true:
- Statement 1 is true: The linear combination \(ay_1 + by_2\) is a solution if \(a + b = 1\).
- Statement 2 is true: The general solution can be expressed as \(y = y_1 + C(y_1 - y_2)\).