Question:medium

Given that the solution of \[ \frac{d^2y}{dx^2} + \alpha \frac{dy}{dx} + \beta y = -e^{-x} \] is \[ y(x) = C_1 e^{-x} + C_2 e^{2x} + x e^{-x}, \] find the values of $\alpha$ and $\beta$.

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If RHS term matches complementary solution, multiply trial solution by $x$. Roots of auxiliary equation directly give coefficients $\alpha$ and $\beta$.
Updated On: Feb 15, 2026
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Solution and Explanation

Step 1: Understanding the Topic
This question involves a second-order linear non-homogeneous differential equation with constant coefficients. We are given the general solution and asked to find the coefficients $\alpha$ and $\beta$ of the differential equation. The key is to work backward from the solution to the original equation.
Step 2: Key Approach - Analyzing the General Solution
The general solution $y(x)$ is composed of two parts: the complementary function $y_c(x)$ (the solution to the homogeneous equation) and a particular solution $y_p(x)$. \[ y(x) = y_c(x) + y_p(x) \] From the given solution, we can identify $y_c(x) = C_1 e^{-x} + C_2 e^{2x}$. The form of $y_c$ is determined by the roots of the auxiliary (characteristic) equation of the homogeneous DE.
Step 3: Detailed Calculation
A. Find the roots of the auxiliary equation:
The terms in the complementary function are $e^{-x}$ and $e^{2x}$. This implies that the roots of the auxiliary equation $m^2 + \alpha m + \beta = 0$ must be: \[ m_1 = -1 \quad \text{and} \quad m_2 = 2 \] B. Construct the auxiliary equation from its roots:
If the roots are -1 and 2, the equation can be written in factored form as: \[ (m - m_1)(m - m_2) = 0 \] \[ (m - (-1))(m - 2) = (m + 1)(m - 2) = 0 \] Expanding this gives: \[ m^2 - 2m + m - 2 = 0 \implies m^2 - m - 2 = 0 \] C. Determine $\alpha$ and $\beta$:
By comparing this equation, $m^2 - m - 2 = 0$, with the standard form $m^2 + \alpha m + \beta = 0$, we can directly identify the coefficients: \[ \alpha = -1 \quad \text{and} \quad \beta = -2 \] D. Verify consistency with the particular solution:
The particular solution given is $y_p(x) = xe^{-x}$. According to the method of undetermined coefficients, since the forcing term $-e^{-x}$ matches a term in the complementary solution, the trial solution for $y_p$ should be of the form $Axe^{-x}$. The given solution is consistent with this rule.
Step 4: Final Answer
The values of the coefficients are $\alpha = -1$ and $\beta = -2$. \[ \boxed{\alpha = -1, \quad \beta = -2} \]
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