Step by step
The question
Question: solve y″ + 2y′ + y = 0, with y(0) = 0 and y′(0) = 4.
The starting values: at x = 0, y is 0 and its slope y′ is 4.
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 + 1 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
3 Solve for r
Two numbers that multiply to 1 and add to 2: 1 and 1, the same number twice.
So it is a square: (r + 1)² = 0.
Only one root: r = −1, twice.
4 Write the solution
One root gives only one solution.
The second one is the same times x.
(Put it in: it works.)
Careful!
That is only one solution twice: (C₁ + C₂)e^(−x).
The second one needs an x.
So y = C₁e^(−x) + C₂xe^(−x).
5 Use the starting values
We need y′ too: the derivative of our y.
Put x = 0 into y: C₁ = 0.
So C₁ is known at once.
Put x = 0 into y′: −C₁ + C₂ = 4.
Put C₁ into ②.
So C₂ = 4.
Put them in: y = 4xe^(−x).
6 Check
Let's check the new part, xe^(−x): its derivatives, by the product rule.
Put them in: everything cancels, 0.
Right!
And the starting values: y(0) = 0 and y′(0) = 4.
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.