Let `y_(1)` and `y_(2)` be two different solutions of the equation
`(dy)/(dx)+P(x).y=Q(x)`. Then `alphay_(1)+betay_(2)` will be solution of the given equation if `alpha + beta=……………….`

Explanation

To solve the problem, we need to find the condition under which the expression \( \alpha y_1 + \beta y_2 \) is a solution of the differential equation

\[ \frac{dy}{dx} + P(x)y = Q(x) \]

given that \( y_1 \) and \( y_2 \) are two different solutions of the same equation.

### Step-by-Step Solution:

1. **Identify the given solutions**:
We know that \( y_1 \) and \( y_2 \) are solutions to the differential equation. Therefore, they satisfy the following equations:

\[ \frac{dy_1}{dx} + P(x)y_1 = Q(x) \quad \text{(1)} \]

\[ \frac{dy_2}{dx} + P(x)y_2 = Q(x) \quad \text{(2)} \]

2. **Formulate the new solution**:
We want to check if \( y = \alpha y_1 + \beta y_2 \) is also a solution of the differential equation.

3. **Differentiate \( y \)**:
Differentiate \( y \) with respect to \( x \):

\[ \frac{dy}{dx} = \alpha \frac{dy_1}{dx} + \beta \frac{dy_2}{dx} \]

4. **Substitute into the differential equation**:
Substitute \( y \) and \( \frac{dy}{dx} \) into the original differential equation:

\[ \alpha \frac{dy_1}{dx} + \beta \frac{dy_2}{dx} + P(x)(\alpha y_1 + \beta y_2) = Q(x) \]

5. **Use the equations (1) and (2)**:
Replace \( \frac{dy_1}{dx} \) and \( \frac{dy_2}{dx} \) using equations (1) and (2):

\[ \alpha (Q(x) - P(x)y_1) + \beta (Q(x) - P(x)y_2) + P(x)(\alpha y_1 + \beta y_2) = Q(x) \]

6. **Simplify the equation**:
Expanding the left-hand side, we get:

\[ \alpha Q(x) - \alpha P(x)y_1 + \beta Q(x) - \beta P(x)y_2 + P(x)(\alpha y_1 + \beta y_2) = Q(x) \]

Combine like terms:

\[ (\alpha + \beta) Q(x) + (-\alpha P(x)y_1 + \alpha P(x)y_1) + (-\beta P(x)y_2 + \beta P(x)y_2) = Q(x) \]

This simplifies to:

\[ (\alpha + \beta) Q(x) = Q(x) \]

7. **Set the coefficients equal**:
For the equation to hold for all \( x \), the coefficients of \( Q(x) \) must be equal:

\[ \alpha + \beta = 1 \]

### Conclusion:
Thus, \( \alpha y_1 + \beta y_2 \) will be a solution of the given differential equation if

\[ \alpha + \beta = 1. \]