Step by step
The question
Question: solve y″ − 2y′ − 15y = 0.
1 Try y = e to the rx
The derivatives of e to the rx are e to the rx again, times r.
So we try y = e^(rx), and look for the right r.
Its derivative: r e^(rx).
Once more: r² e^(rx).
2 The characteristic equation
Put them into the equation.
Every term has e^(rx): take it out.
e to a power is never 0.
So the bracket must be 0.
r² − 2r − 15 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
3 Solve for r
Two numbers that multiply to −15 and add to −2: 3 and −5.
So it factors: (r + 3)(r − 5) = 0.
Each bracket can be 0: r = −3 or r = 5.
Careful!
(r + 3) is 0 when r is −3: the opposite sign of the number in the bracket.
4 Write the solution
Each root gives a solution e^(rx).
Any mix of the two works, so we add them with two numbers, C₁ and C₂.
So y = C₁e^(−3x) + C₂e^(5x).
5 Check
Let's check: each root must make r² − 2r − 15 = 0.
r = −3 gives 0.
And r = 5 gives 0.
Right!
Let's look back.
First: try y = e to the rx.
Then: the characteristic equation.
Then: solve for r.
Then: write the solution.
Last: check.