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What is the solution to the equation y'' + 4y = 0?

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Question: What is the solution to the equation y\'\' + 4y = 0?

Options:

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

Correct Answer: y = C1 cos(2x) + C2 sin(2x)

Solution: 

The characteristic equation is r^2 + 4 = 0, giving complex roots. The general solution is y = C1 cos(2x) + C2 sin(2x).

What is the solution to the equation y'' + 4y = 0?

Practice Questions

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

Questions & Step-by-Step Solutions

What is the solution to the equation y'' + 4y = 0?
  • Step 1: Start with the given equation: y'' + 4y = 0.
  • Step 2: Identify the characteristic equation by replacing y'' with r^2 and y with 1: r^2 + 4 = 0.
  • Step 3: Solve the characteristic equation for r: r^2 = -4.
  • Step 4: Take the square root of both sides: r = ±2i (where i is the imaginary unit).
  • Step 5: Since the roots are complex (±2i), use the formula for the general solution of second-order differential equations with complex roots.
  • Step 6: The general solution is: y = C1 cos(2x) + C2 sin(2x), where C1 and C2 are constants.
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