Question

Consider the differential equation \[ \frac{d^{2}y}{dx^{2}} + 8\frac{dy}{dx} + 16y = 0 \] and the boundary conditions \( y(0) = 1 \) and \( \frac{dy}{dx}(0) = 0 \). The solution to this equation is:

• A. \( y = (1 + 2x)e^{-4x} \)
• B. \( y = (1 - 4x)e^{-4x} \)
• C. \( y = (1 + 8x)e^{-4x} \)
• D. \( y = (1 + 4x)e^{-4x} \)

Hint:

Repeated roots in differential equations yield solutions of the form \( (C_{1} + C_{2}x)e^{mx} \).

Answer 1

The Correct Option is D

Step 1: Solve the characteristic equation. 
The ODE \[ y'' + 8y' + 16y = 0 \] gives the characteristic equation \[ m^{2} + 8m + 16 = 0 = (m+4)^{2}. \] Thus, the repeated root is \( m = -4 \). 
 

Step 2: General solution for repeated roots. 
\[ y = (C_{1} + C_{2}x)e^{-4x}. \] 
 

Step 3: Apply boundary conditions. 
From \( y(0) = 1 \), we get \( C_{1} = 1 \). 
Differentiate: \[ y' = C_{2}e^{-4x} - 4(C_{1} + C_{2}x)e^{-4x}. \] At \( x = 0 \): \[ 0 = y'(0) = C_{2} - 4C_{1} \Rightarrow C_{2} = 4. \] 
 

Step 4: Final solution. 
\[ y = (1 + 4x)e^{-4x}. \]