Apply two initial conditions to distinct real roots
After obtaining a complementary solution, use both the displacement and derivative conditions to find the constants. Differentiate the complete expression before substituting the initial point. A sign error in the derivative of a decaying exponential changes the constants even when the characteristic equation was solved correctly.
Worked practice example
Solve y″ − y′ − 2y = 0 with y(0) = 1 and y′(0) = 0.
| Step | Working |
|---|---|
| Find the roots | r² − r − 2 = (r − 2)(r + 1), giving r = 2 and −1. |
| Write and differentiate | y = C₁e^(2x) + C₂e^(−x); y′ = 2C₁e^(2x) − C₂e^(−x). |
| Apply both conditions | C₁ + C₂ = 1 and 2C₁ − C₂ = 0. |
| Solve the constants | C₁ = 1/3 and C₂ = 2/3. Thus y = e^(2x)/3 + 2e^(−x)/3. |
Check your method
At x = 0, the coefficients sum to 1 and the derivative is 2/3 − 2/3 = 0. The growing exponential eventually dominates, so a solution that decays to zero for this problem would need rechecking. A nonzero coefficient on a positive characteristic root produces growth.
Try it yourself
Solve the same differential equation with y(0) = 0 and y′(0) = 3.
Show the answer and method
The equations are C₁ + C₂ = 0 and 2C₁ − C₂ = 3. Hence C₁ = 1, C₂ = −1 and y = e^(2x) − e^(−x).
This written practice example was prepared for Jemshah revision. It is additional practice, not an official examination question or a transcript of the video.