Question:

The solution of the differential equation \( \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} + 2y = 0 \) is given by

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When solving second-order linear differential equations with constant coefficients, always find the characteristic equation by assuming a solution of the form \( y = e^{rx} \), then solve for the roots of the characteristic equation.
Updated On: Feb 3, 2026
  • \( y = C_1 e^x + C_2 e^{2x} \)
  • \( y = C_1 e^x + C_2 e^{-2x} \)
  • \( y = C_1 e^x + C_2 e^{2x} \)
  • \( y = C_1 e^x + C_2 e^{2x} \)
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The Correct Option is C

Solution and Explanation

Step 1: Understanding the differential equation.
We are given the second-order linear homogeneous differential equation: \[ \frac{d^2 y}{dx^2} - 3 \frac{dy}{dx} + 2y = 0 \] This is a linear differential equation with constant coefficients. To solve this type of equation, we first write down the characteristic equation and solve for the roots.
Step 2: Finding the characteristic equation.
Assume that the solution is of the form \( y = e^{rx} \), where \( r \) is a constant. Then, we can calculate the first and second derivatives: \[ \frac{dy}{dx} = r e^{rx}, \quad \frac{d^2 y}{dx^2} = r^2 e^{rx} \] Substituting these into the given equation: \[ r^2 e^{rx} - 3r e^{rx} + 2 e^{rx} = 0 \] Factor out \( e^{rx} \) (which is never zero): \[ e^{rx} (r^2 - 3r + 2) = 0 \] Thus, the characteristic equation is: \[ r^2 - 3r + 2 = 0 \]
Step 3: Solving the characteristic equation.
Now, solve the quadratic equation \( r^2 - 3r + 2 = 0 \). Factorizing: \[ (r - 1)(r - 2) = 0 \] So, the roots are: \[ r_1 = 1 \quad \text{and} \quad r_2 = 2 \]
Step 4: General solution.
Since the roots \( r_1 = 1 \) and \( r_2 = 2 \) are real and distinct, the general solution to the differential equation is: \[ y(x) = C_1 e^{r_1 x} + C_2 e^{r_2 x} \] Substituting the values of \( r_1 \) and \( r_2 \): \[ y(x) = C_1 e^x + C_2 e^{2x} \]
Step 5: Conclusion.
The general solution to the given differential equation is: \[ y(x) = C_1 e^x + C_2 e^{2x} \] Therefore, the correct answer is: \[ \boxed{\text{(C) } y = C_1 e^x + C_2 e^{2x}} \]
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