Step by step
The question
Question: solve y″ + 2y′ + 5y = 0, with y(0) = 1 and y′(0) = 3.
The starting values: at x = 0, y is 1 and its slope y′ is 3.
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 + 5 = 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 = 2, c = 5.
Put the numbers in.
Under the root: −16.
A minus under the root: √(−16) = 4i, because i = √(−1).
Divide both parts by 2: r = −1 ± 2i.
Two complex roots.
Careful!
Don't lose the real part, −1.
It gives e^(−x) in front.
4 Write the solution
Complex roots give cosines and sines.
The real part −1 goes in the e, the 2 goes inside cos and sin.
So y = e^(−x)(C₁cos(2x) + C₂sin(2x)).
5 Use the starting values
We need y′ too: the derivative of our y.
Put x = 0 into y: C₁ = 1.
So C₁ is known at once.
Put x = 0 into y′: −C₁ + 2C₂ = 3.
Put C₁ into ②.
So C₂ = 2.
Put them in: y = e^(−x)(cos(2x) + 2sin(2x)).
6 Check
Let's check one root, −1 + 2i.
Square it, with i² = −1: −3 − 4i.
Into r² + 2r + 5: the numbers cancel and the i parts cancel.
0.
Right!
And the starting values: y(0) = 1 and y′(0) = 3.
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.