Step by step
The question
Question: solve 2y″ + 3y′ + y = 0, with y(0) = 1 and y′(0) = 0.
The starting values: at x = 0, y is 1 and its slope y′ is 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.
2r² + 3r + 1 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
3 Solve for r
It does not factor nicely, so we use the formula, with a = 2, b = 3, c = 1.
Put the numbers in.
Under the root: 1.
Minus gives −1, plus gives −1/2.
Careful!
The whole top is over 2a, not only the root.
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^(−x) + C₂e^(−x/2).
5 Use the starting values
We need y′ too: the derivative of our y.
Put x = 0 into y: C₁ + C₂ = 1.
Put x = 0 into y′: −C₁ − ½C₂ = 0.
From ①: C₁ = 1 − C₂.
Put it into ②.
Open the bracket and collect: −1 × 1 moves over.
So C₂ = 2, and C₁ = −1.
Put them in: y = −e^(−x) + 2e^(−x/2).
6 Check
Let's check: each root must make 2r² + 3r + 1 = 0.
r = −1 gives 0.
And r = −1/2 gives 0.
Right!
And the starting values: y(0) = 1 and y′(0) = 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.
Then: use the starting values.
Last: check.