Question

Consider the following ordinary differential equation: \[ 4 \frac{d^{2} y}{dx^{2}} - 4 \frac{dy}{dx} + y = 0. \] Given that \( c_{1} \) and \( c_{2} \) are constants, the general solution of the differential equation is

• A. \( y = (c_{1} + c_{2} x)e^{x} \)
• B. \( y = c_{1} e^{x/2} + c_{2} e^{x} \)
• C. \( y = c_{1} e^{x} + c_{2} e^{2x} \)
• D. \( y = (c_{1} + c_{2} x)e^{x/2} \)

Hint:

Repeated roots of the characteristic equation always produce solutions of the form \((c_{1} + c_{2}x)e^{rx}\).

Answer 1

The Correct Option is D

To solve the differential equation \[ 4y'' - 4y' + y = 0, \] we write its characteristic equation: \[ 4r^{2} - 4r + 1 = 0. \]

Step 1: Solve the characteristic equation.
The discriminant is \[ (-4)^{2} - 4(4)(1) = 16 - 16 = 0, \] so the equation has a repeated real root. Solving, \[ r = \frac{4}{8} = \frac{1}{2}. \] Thus, the repeated root is \( r = \frac{1}{2} \).

Step 2: Write the general solution for a repeated root.
If the root \( r \) is repeated, the solution is: \[ y = (c_{1} + c_{2} x)e^{rx}. \] Substituting \( r = \tfrac{1}{2} \): \[ y = (c_{1} + c_{2} x)e^{x/2}. \]

Step 3: Match with options.
The expression matches option (D).

Final Answer: \( (c_{1} + c_{2} x)e^{x/2} \)