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What is the general solution of the equation y'' - 4y' + 4y = 0?

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Question: What is the general solution of the equation y\'\' - 4y\' + 4y = 0?

Options:

  1. y = (C1 + C2x)e^(2x)
  2. y = C1 e^(2x) + C2 e^(-2x)
  3. y = C1 e^(4x) + C2 e^(-4x)
  4. y = C1 cos(2x) + C2 sin(2x)

Correct Answer: y = (C1 + C2x)e^(2x)

Solution: 

The characteristic equation has a repeated root r = 2. The general solution is y = (C1 + C2x)e^(2x).

What is the general solution of the equation y'' - 4y' + 4y = 0?

Practice Questions

Q1
What is the general solution of the equation y'' - 4y' + 4y = 0?
  1. y = (C1 + C2x)e^(2x)
  2. y = C1 e^(2x) + C2 e^(-2x)
  3. y = C1 e^(4x) + C2 e^(-4x)
  4. y = C1 cos(2x) + C2 sin(2x)

Questions & Step-by-Step Solutions

What is the general solution of the equation y'' - 4y' + 4y = 0?
  • Step 1: Write down the given differential equation: y'' - 4y' + 4y = 0.
  • Step 2: Identify the characteristic equation by replacing y'' with r^2, y' with r, and y with 1: r^2 - 4r + 4 = 0.
  • Step 3: Factor the characteristic equation: (r - 2)(r - 2) = 0.
  • Step 4: Solve for r: The repeated root is r = 2.
  • Step 5: Write the general solution for a repeated root: y = (C1 + C2x)e^(rx).
  • Step 6: Substitute r = 2 into the general solution: y = (C1 + C2x)e^(2x).
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