Choose a complementary solution from the characteristic roots
For ay″ + by′ + cy = 0 with constant coefficients and a ≠ 0, try y = e^(rx). The characteristic equation is ar² + br + c = 0. Two distinct real roots give two different exponentials. A repeated real root needs x multiplying one solution. Complex roots α ± iβ give an exponentially weighted sine and cosine. A second-order general solution normally contains two independent constants.
Worked practice example
Solve y″ + 4y′ + 13y = 0.
| Step | Working |
|---|---|
| Characteristic equation | r² + 4r + 13 = 0. |
| Complete the square | (r + 2)² + 9 = 0, so r = −2 ± 3i. |
| Real general solution | y = e^(−2x)[C₁ cos(3x) + C₂ sin(3x)]. |
| Interpretation | The factor e^(−2x) decays for increasing x, while sine and cosine describe oscillation. |
Check your method
For a repeated root r, use (C₁ + C₂x)e^(rx), not two copies of e^(rx). For distinct real roots r₁,r₂ use C₁e^(r₁x) + C₂e^(r₂x). If there is a nonzero right-hand side, this complementary solution must be combined with a particular solution.
Try it yourself
Write the complementary solution of y″ − 6y′ + 9y = 0.
Show the answer and method
The characteristic polynomial is (r − 3)². The repeated-root solution is y = (C₁ + C₂x)e^(3x).
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.