Let `y_(1) and y_(2)` be the solutions of the differential equation `(dy)/(dx)+Py=Q`, where P and Q are functional of x.
Statement-1 : `(y_(2)-y_(1))/(y_(1))=Ce^(-int(Q)/(y1)dx)`
Statement-2 : If `y_(2)=y_(1)z`, then `z=1+Ce^(int(-Q)/(y1)dx)`, where C is an arbitrary constant.

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 problem, we need to analyze the given differential equation and the statements provided. The differential equation is given as:

\[ \frac{dy}{dx} + Py = Q \]

where \( P \) and \( Q \) are functions of \( x \).

### Step 1: Understand the Solutions
Let \( y_1 \) and \( y_2 \) be two solutions of the differential equation. We can express \( y_2 \) in terms of \( y_1 \) as:

\[ y_2 = y_1 z \]

where \( z \) is a function of \( x \).

### Step 2: Differentiate \( y_2 \)
Using the product rule, we differentiate \( y_2 \):

\[ \frac{dy_2}{dx} = \frac{d(y_1 z)}{dx} = \frac{dy_1}{dx} z + y_1 \frac{dz}{dx} \]

### Step 3: Substitute into the Differential Equation
Substituting \( y_1 \) and \( y_2 \) into the original differential equation:

For \( y_1 \):
\[ \frac{dy_1}{dx} + Py_1 = Q \]

For \( y_2 \):
\[ \frac{dy_2}{dx} + Py_2 = Q \]

Substituting \( y_2 = y_1 z \) into the second equation gives:

\[ \frac{dy_1}{dx} z + y_1 \frac{dz}{dx} + P(y_1 z) = Q \]

### Step 4: Rearranging the Equation
Now, we can rearrange the equation:

\[ \frac{dy_1}{dx} z + Py_1 z + y_1 \frac{dz}{dx} = Q \]

Since \( \frac{dy_1}{dx} + Py_1 = Q \), we can substitute \( Q \) from the first equation:

\[ Q z + y_1 \frac{dz}{dx} = Q \]

### Step 5: Isolate \( \frac{dz}{dx} \)
Rearranging gives:

\[ y_1 \frac{dz}{dx} = Q(1 - z) \]

Dividing both sides by \( y_1 \):

\[ \frac{dz}{dx} = \frac{Q(1 - z)}{y_1} \]

### Step 6: Solving the Differential Equation
This is a separable differential equation. We can separate variables:

\[ \frac{dz}{1 - z} = \frac{Q}{y_1} dx \]

Integrating both sides gives:

\[ -\ln|1 - z| = \int \frac{Q}{y_1} dx + C \]

### Step 7: Exponentiate to Solve for \( z \)
Exponentiating both sides results in:

\[ 1 - z = Ce^{-\int \frac{Q}{y_1} dx} \]

Thus,

\[ z = 1 - Ce^{-\int \frac{Q}{y_1} dx} \]

### Step 8: Relating \( y_2 \) and \( y_1 \)
Since \( y_2 = y_1 z \):

\[ y_2 = y_1 \left(1 + Ce^{-\int \frac{Q}{y_1} dx}\right) \]

### Step 9: Final Formulation
Now, we can express \( y_2 - y_1 \):

\[ y_2 - y_1 = y_1 \left(1 + Ce^{-\int \frac{Q}{y_1} dx}\right) - y_1 = y_1 Ce^{-\int \frac{Q}{y_1} dx} \]

Dividing by \( y_1 \):

\[ \frac{y_2 - y_1}{y_1} = Ce^{-\int \frac{Q}{y_1} dx} \]

### Conclusion
Both statements are true based on our derivation.