Step by step
The question
Question: solve 4y″ + 4y′ + y = 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.
4r² + 4r + 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 = 4, b = 4, c = 1.
Put the numbers in.
Under the root: 0.
Under the root is 0, so plus and minus give the same: r = −1/2, 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/2).
The second one needs an x.
So y = C₁e^(−x/2) + C₂xe^(−x/2).
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′: −½C₁ + C₂ = 1.
Put C₁ into ②.
So C₂ = 2.
Put them in: y = 2e^(−x/2) + 2xe^(−x/2).
6 Check
Let's check the new part, xe^(−x/2): its derivatives, by the product rule.
Put them in: everything cancels, 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.