Question

Solution of the differential equation $y'' + y' + 0.25y = 0$ with the initial values $y(0)=3.0$ and $y'(0)=-3.5$ is

• A. $y = (3 - 2x)e^{0.5x}$
• B. $y = (3 - 2x)e^{-0.25x}$
• C. $y = (3 - 2x)e^{-0.5x}$
• D. $y = (2 - 3x)e^{-0.5x}$

Hint:

Repeated real roots produce solutions of the form $(C_1 + C_2 x)e^{rx}$.

Answer 1

The Correct Option is C

Step 1: Solve characteristic equation.
\[ r^2 + r + 0.25 = 0 \] \[ r = \frac{-1 \pm \sqrt{1 - 1}}{2} = -0.5 \] Repeated real root → solution is \[ y = (C_1 + C_2 x)e^{-0.5x}. \]

Step 2: Apply initial conditions.
From $y(0)=3$: \[ C_1 = 3. \] Differentiate: \[ y' = C_2 e^{-0.5x} - 0.5(C_1 + C_2 x)e^{-0.5x}. \] At $x=0$, \[ y'(0)= C_2 - 0.5C_1 = -3.5. \] \[ C_2 - 1.5 = -3.5 $\Rightarrow$ C_2 = -2. \] Final solution: \[ y = (3 - 2x)e^{-0.5x}. \] Matches Option (C).