Question:

To determine the correct form of the general solution for the given differential equation, let's proceed step by step. Step 1: Understand the given differential equation The differential equation given is $\frac{dy}{dx} + P(x)y = Q(x)$. This is a first-order linear differential equation. Step 2: Recall the general form of the solution for a first-order linear differential equation For the homogeneous part of the equation, $\frac{dy}{dx} + P(x)y = 0$, the solution can be found using an integrating factor. However, when we have particular solutions $y_1$ and $y_2$, we can use them to construct the general solution. Step 3: Consider the given particular solutions $y_1$ and $y_2$ Given $y_1$ and $y_2$ are solutions to the differential equation, we have: $$ \begin{aligned} \frac{dy_1}{dx} + P(x)y_1 &= Q(x) \\ \frac{dy_2}{dx} + P(x)y_2 &= Q(x) \end{aligned} $$ Step 4: Examine the nature of the general solution For a linear differential equation, if $y_1$ and $y_2$ are solutions, then any linear combination of $y_1$ and $y_2$ is also a solution. However, the question hints at a specific form involving $y_1$ and $y_2$. Step 5: Evaluate the given options Looking at the options provided: 1. $y = y_1 + c(y_1 - y_2)$ 2. $y = y_1 + c(y_1 + y_2)$ 3. $\alpha y_1 + \beta y_2$ is a solution of $\alpha + \beta = 1$ 4. $\alpha y_1 + \beta y_2$ is a solution of $\alpha - \beta = 1$ We need to determine which of these represents the general solution or a valid solution form for the given differential equation. Step 6: Analyze option 1 Option 1 suggests $y = y_1 + c(y_1 - y_2)$. Let's verify if this form satisfies the differential equation. Step 7: Analyze option 3 Option 3 suggests $\alpha y_1 + \beta y_2$ is a solution when $\alpha + \beta = 1$. This implies a linear combination of $y_1$ and $y_2$ with coefficients summing to 1. Step 8: Determine the validity of each option Given that $y_1$ and $y_2$ are solutions, the general solution should indeed involve a linear combination of these solutions. The condition $\alpha + \beta = 1$ essentially normalizes the solutions, suggesting that any solution can be represented as a weighted sum of $y_1$ and $y_2$ where the weights sum to 1. Step 9: Conclusion on the general solution form The general solution for a first-order linear differential equation, given two particular solutions $y_1$ and $y_2$, can indeed be expressed as a linear combination of these solutions. The form $y = y_1 + c(y_1 - y_2)$ seems to misinterpret the principle of linear combinations in this context, as it does not directly represent a linear combination of $y_1$ and $y_2$ with arbitrary coefficients. Instead, $\alpha y_1 + \beta y_2$ with $\alpha + \beta = 1$ suggests a specific linear combination that always results in a valid solution, aligning with the principle that any linear combination of solutions to a linear differential equation is also a solution. Therefore: The correct options are 1 and 3, as they represent valid forms for solutions of the given differential equation, with option 1 being a general form and option 3 specifying a condition under which a linear combination of $y_1$ and $y_2$ is a solution. <div class="key-concept"><strong>Key Concept:</strong> The solution space of a linear ODE is a vector space; any linear combination of solutions with coefficients summing to 1 is also a solution.</div> <div class="trap-box"><strong>Trap:</strong> Students may forget that the coefficient constraint $\alpha + \beta = 1$ is necessary for linear combinations to be solutions to the homogeneous equation.</div>