Question

Consider the differential equation
\[ y'' - 4y' + 20y = 0 \] with $y\left(\frac{\pi}{2}\right) = 0$, and $y'\left(\frac{\pi}{2}\right) = 1$, then the value of $y\left(\frac{\pi}{8}\right)$ is

• A. $-\frac{1}{4} e^{-3\pi/8}$
• B. $-\frac{1}{4} e^{3\pi/8}$
• C. $-\frac{1}{4} e^{3\pi/4}$
• D. $-\frac{1}{4} e^{-3\pi/4}$

Hint:

Always double check the angle evaluated in trig functions! At $x = \pi/8$, $4x = \pi/2$, which simplifies $\sin(4x) = 1$. This eliminates trigonometric factors cleanly!

Answer 1

The Correct Option is D

Step 1: Concept
This is a second-order linear homogeneous differential equation with constant coefficients. We solve it using the auxiliary equation method, apply the initial boundary conditions, and evaluate at the target point.

Step 2: Key Formulas and Approach
1. For $a y'' + b y' + c y = 0$, write auxiliary equation $a m^2 + b m + c = 0$. 2. If roots are complex $\alpha \pm i \beta$, general solution is: \[ y(x) = e^{\alpha x} \left( C_1 \cos(\beta x) + C_2 \sin(\beta x) \right) \]

Step 3: Step-by-step Explanation

• Write the auxiliary equation for $y'' - 4y' + 20y = 0$: \[ m^2 - 4m + 20 = 0 \]
• Solve for $m$: \[ m = \frac{4 \pm \sqrt{16 - 80}}{2} = \frac{4 \pm \sqrt{-64}}{2} = 2 \pm 4i \] Here $\alpha = 2$ and $\beta = 4$.
• Write the general solution: \[ y(x) = e^{2x} \left( C_1 \cos(4x) + C_2 \sin(4x) \right) \]
• Apply initial condition $y\left(\frac{\pi}{2}\right) = 0$: \[ y\left(\frac{\pi}{2}\right) = e^{\pi} \left( C_1 \cos(2\pi) + C_2 \sin(2\pi) \right) = e^{\pi} (C_1 (1) + C_2 (0)) = C_1 e^{\pi} = 0 \] Since $e^{\pi} \neq 0$, we get $C_1 = 0$.
• So $y(x) = C_2 e^{2x} \sin(4x)$.
• Compute $y'(x)$: \[ y'(x) = C_2 \left[ 2e^{2x} \sin(4x) + 4e^{2x} \cos(4x) \right] \]
• Apply initial condition $y'\left(\frac{\pi}{2}\right) = 1$: \[ y'\left(\frac{\pi}{2}\right) = C_2 e^{\pi} \left[ 2(0) + 4(1) \right] = 4 C_2 e^{\pi} = 1 \implies C_2 = \frac{1}{4} e^{-\pi} \]
• Write the explicit solution: \[ y(x) = \frac{1}{4} e^{-\pi} e^{2x} \sin(4x) = \frac{1}{4} e^{2x - \pi} \sin(4x) \]
• Evaluate at $x = \frac{\pi}{8}$: \[ 4x = 4\left(\frac{\pi}{8}\right) = \frac{\pi}{2} \implies \sin\left(\frac{\pi}{2}\right) = 1 \] \[ 2x - \pi = 2\left(\frac{\pi}{8}\right) - \pi = \frac{\pi}{4} - \pi = -\frac{3\pi}{4} \] \[ y\left(\frac{\pi}{8}\right) = \frac{1}{4} e^{-3\pi/4} (1) = \frac{1}{4} e^{-3\pi/4} \]

Step 4: Final Answer
The solution evaluated at $x = \pi/8$ is $-\frac{1}{4} e^{-3\pi/4}$. Thus, Option (D) is correct.