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What is the solution to the equation y'' - 3y' + 2y = 0?

Practice Questions

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

Questions & Step-by-Step Solutions

What is the solution to the equation y'' - 3y' + 2y = 0?
  • Step 1: Identify the given differential equation, which is y'' - 3y' + 2y = 0.
  • Step 2: Write the characteristic equation by replacing y'' with r^2, y' with r, and y with 1. This gives us r^2 - 3r + 2 = 0.
  • Step 3: Factor the characteristic equation r^2 - 3r + 2. It factors to (r - 1)(r - 2) = 0.
  • Step 4: Solve for r by setting each factor equal to zero. This gives us r - 1 = 0 (so r = 1) and r - 2 = 0 (so r = 2).
  • Step 5: Write the general solution using the roots found. The general solution is y = C1 e^(1x) + C2 e^(2x), where C1 and C2 are constants.
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