Step by step
The question
Question: solve y″ − 3y′ + 2y = 0, with y(0) = 0 and y′(0) = 1.
The starting values: at x = 0, y is 0 and its slope y′ is 1.
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 − 3r + 2 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
3 Solve for r
Two numbers that multiply to 2 and add to −3: −1 and −2.
So it factors: (r − 1)(r − 2) = 0.
Each bracket can be 0: r = 1 or r = 2.
Careful!
(r − 1) is 0 when r is 1: 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 = C₁eˣ + C₂e^(2x).
5 Use the starting values
We need y′ too: the derivative of our y.
Put x = 0 into y: C₁ + C₂ = 0.
Put x = 0 into y′: C₁ + 2C₂ = 1.
From ①: C₁ = 0 − C₂.
Put it into ②.
Open the bracket and collect: 1 × 0 moves over.
So C₂ = 1, and C₁ = −1.
Put them in: y = − eˣ + e^(2x).
6 Check
Let's check: each root must make r² − 3r + 2 = 0.
r = 1 gives 0.
And r = 2 gives 0.
Right!
And the starting values: y(0) = 0 and y′(0) = 1.
Right!
Let's look back.
First: try y = e to the rx.
Then: the characteristic equation.
Then: solve for r.
Then: write the solution.
Then: use the starting values.
Last: check.