Step by step
The question
Question: solve y″ + 4y′ + 13y = 0, with y(0) = 2 and y′(0) = −1.
The starting values: at x = 0, y is 2 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 + 4r + 13 = 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 = 1, b = 4, c = 13.
Put the numbers in.
Under the root: −36.
A minus under the root: √(−36) = 6i, because i = √(−1).
Divide both parts by 2: r = −2 ± 3i.
Two complex roots.
Careful!
Don't lose the real part, −2.
It gives e^(−2x) in front.
4 Write the solution
Complex roots give cosines and sines.
The real part −2 goes in the e, the 3 goes inside cos and sin.
So y = e^(−2x)(C₁cos(3x) + C₂sin(3x)).
5 Use the starting values
We need y′ too: the derivative of our y.
Put x = 0 into y: C₁ = 2.
So C₁ is known at once.
Put x = 0 into y′: −2C₁ + 3C₂ = −1.
Put C₁ into ②.
So C₂ = 1.
Put them in: y = e^(−2x)(2cos(3x) + sin(3x)).
6 Check
Let's check one root, −2 + 3i.
Square it, with i² = −1: −5 − 12i.
Into r² + 4r + 13: the numbers cancel and the i parts cancel.
0.
Right!
And the starting values: y(0) = 2 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.