Step by step
The question
Question: solve y″ + 2y′ − 8y = 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 − 8 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
3 Solve for r
Two numbers that multiply to −8 and add to 2: 4 and −2.
So it factors: (r + 4)(r − 2) = 0.
Each bracket can be 0: r = −4 or r = 2.
Careful!
(r + 4) is 0 when r is −4: 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^(−4x) + C₂e^(2x).
5 Check
Let's check: each root must make r² + 2r − 8 = 0.
r = −4 gives 0.
And r = 2 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.