Question

The solution of \( y'' - y' - 2y = 0 \) is

• A. \( e^{2x} + e^{-2x} \)
• B. \( c_1 e^{2x} + c_2 e^{-2x} \)
• C. \( c_1 e^{2x} + c_2 e^{-x} \)
• D. \( e^{2x} - c_2 e^{-2x} \)

Hint:

For a second-order linear homogeneous differential equation \( ay'' + by' + cy = 0 \), the characteristic equation is \( ar^2 + br + c = 0 \). Distinct real roots \( r_1, r_2 \) give the solution \( y = c_1 e^{r_1 x} + c_2 e^{r_2 x} \).

Answer 1

The Correct Option is C

Step 1: Solve the differential equation.
The given equation is \( y'' - y' - 2y = 0 \), a second-order linear homogeneous differential equation with constant coefficients. Assume a solution of the form \( y = e^{rx} \). The characteristic equation is: \[ r^2 - r - 2 = 0. \] Solve the quadratic: \[ r = \frac{1 \pm \sqrt{1 + 8}}{2} = \frac{1 \pm \sqrt{9}}{2} = \frac{1 \pm 3}{2}, \] \[ r = 2 \quad \text{or} \quad r = -1. \] Since the roots are distinct, the general solution is: \[ y = c_1 e^{r_1 x} + c_2 e^{r_2 x} = c_1 e^{2x} + c_2 e^{-x}, \] where \( c_1 \) and \( c_2 \) are arbitrary constants. Step 2: Evaluate the options.
(1) \( e^{2x} + e^{-2x} \): Incorrect, as the second exponent should be \( e^{-x} \), not \( e^{-2x} \), and it lacks arbitrary constants. Incorrect.
(2) \( c_1 e^{2x} + c_2 e^{-2x} \): Incorrect, as the second exponent should be \( e^{-x} \), not \( e^{-2x} \). Incorrect.
(3) \( c_1 e^{2x} + c_2 e^{-x} \): Correct, matches the general solution. Correct.
(4) \( e^{2x} - c_2 e^{-2x} \): Incorrect, as the second exponent is wrong, and the first term lacks a constant coefficient. Incorrect.
Step 3: Select the correct answer.
The solution is \( c_1 e^{2x} + c_2 e^{-x} \), matching option (3).