Handle a repeated root without losing a solution
When a characteristic root repeats, two copies of the same exponential are dependent and cannot represent every initial condition. The second independent solution is x times that exponential. The product rule is needed when differentiating this solution. Repeated negative roots give a nonoscillatory response that still tends to zero as the independent variable increases.
Worked practice example
Solve y″ + 6y′ + 9y = 0 with y(0) = 2 and y′(0) = −1.
| Step | Working |
|---|---|
| Repeated root | r² + 6r + 9 = (r + 3)², so r = −3. |
| General solution | y = (C₁ + C₂x)e^(−3x). |
| Differentiate | y′ = [C₂ − 3C₁ − 3C₂x]e^(−3x). |
| Use the initial data | C₁ = 2 and C₂ − 6 = −1, so C₂ = 5; y = (2 + 5x)e^(−3x). |
Check your method
The derivative at zero is 5 − 6 = −1. Keeping the x term is essential even though x = 0 in the initial condition: its derivative supplies the second independent constant. In mechanical models this repeated-root structure appears at critical damping when the model’s mass, damping and stiffness give a repeated root.
Try it yourself
Solve y″ + 4y′ + 4y = 0 with y(0) = 1 and y′(0) = 0.
Show the answer and method
The root is −2 twice, so y = (C₁ + C₂x)e^(−2x). C₁ = 1 and C₂ − 2C₁ = 0 give y = (1 + 2x)e^(−2x).
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.