Translate derivative notation into an operator polynomial
Let D mean differentiation with respect to x: Dy = y′ and D²y = y″. A constant-coefficient equation can then be written F(D)y = f(x). This notation organises factorisation and complementary solutions. It does not turn differentiation into ordinary multiplication by a number; an operator acts on the function following it. When coefficients vary with x, the order of multiplication and differentiation matters.
Worked practice example
Solve (D² − 5D + 6)y = 0.
| Step | Working |
|---|---|
| Translate | The equation is y″ − 5y′ + 6y = 0. |
| Factor the polynomial | F(r) = r² − 5r + 6 = (r − 2)(r − 3). |
| Find the independent solutions | e^(2x) and e^(3x). |
| Combine | y = C₁e^(2x) + C₂e^(3x). |
Check your method
For a constant a, D(e^(ax)) = ae^(ax), so F(D)e^(ax) = F(a)e^(ax). That is why the characteristic roots appear. As a separate check on operator order, D(xy) = xy′ + y, whereas xDy = xy′; they differ by y. The simple polynomial factorisation here is valid because the coefficients are constant.
Try it yourself
Expand (D + 1)(D + 2)y and state its complementary solution.
Show the answer and method
The expanded equation is y″ + 3y′ + 2y = 0. Its roots are −1 and −2, so y = C₁e^(−x) + C₂e^(−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.