Find the general solution of the differential equation y'' - 5y' + 6y = 0.

Practice Questions

Q1
Find the general solution of the differential equation y'' - 5y' + 6y = 0.
  1. y = C1 e^(2x) + C2 e^(3x)
  2. y = C1 e^(3x) + C2 e^(2x)
  3. y = C1 e^(x) + C2 e^(2x)
  4. y = C1 e^(4x) + C2 e^(5x)

Questions & Step-by-Step Solutions

Find the general solution of the differential equation y'' - 5y' + 6y = 0.
Correct Answer: y = C1 e^(2x) + C2 e^(3x)
  • Step 1: Write down the given differential equation: y'' - 5y' + 6y = 0.
  • Step 2: Identify the characteristic equation by replacing y'' with r^2, y' with r, and y with 1. This gives us r^2 - 5r + 6 = 0.
  • Step 3: Factor the characteristic equation r^2 - 5r + 6. Look for two numbers that multiply to 6 and add to -5. The numbers are -2 and -3.
  • Step 4: Write the factored form: (r - 2)(r - 3) = 0.
  • Step 5: Set each factor equal to zero to find the roots: r - 2 = 0 gives r = 2, and r - 3 = 0 gives r = 3.
  • Step 6: Since we have two distinct real roots (2 and 3), the general solution of the differential equation is: y = C1 e^(2x) + C2 e^(3x), where C1 and C2 are constants.
  • Homogeneous Linear Differential Equations – The question tests the ability to solve a second-order linear homogeneous differential equation with constant coefficients.
  • Characteristic Equation – It assesses the understanding of deriving the characteristic equation from the differential equation and finding its roots.
  • General Solution – The question evaluates the ability to write the general solution based on the roots of the characteristic equation.
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