Steply

Solve y'' + 4y = 0

Step by step, with the reason for every step · Second order · Differential Equations

Step by step

The question

Question: solve y″ + 4y = 0.

y″ + 4y = 0
second order, linear, constant numbers, 0 on the right

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.

y = e^(rx): each derivative keeps the same shape
y = e^(rx)

Its derivative: r e^(rx).

y′ = re^(rx)

Once more: r² e^(rx).

y″ = r2e^(rx)

2 The characteristic equation

Put them into the equation.

r2e^(rx) + 4e^(rx) = 0

Every term has e^(rx): take it out.

e^(rx)(r2 + 4) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

r² + 4 = 0: the characteristic equation.

y″ became r², y′ became r, y became 1.

r2 + 4 = 0

3 Solve for r

It does not factor nicely, so we use the formula, with a = 1, b = 0, c = 4.

r = −b ± √b² − 4ac2a

Put the numbers in.

r = 0 ± √0² − 4 × 1 × 42 × 1

Under the root: −16.

r = 0 ± √−162

A minus under the root: √(−16) = 4i, because i = √(−1).

√−16 = √16 × √−1 = 4i
r = 0 ± 4i2

Divide both parts by 2: r = ±2i.

Two complex roots.

r = ±2i

Careful!

Use the root, 2, not the number under it.

Careful: not cos(4x) ✗

4 Write the solution

Complex roots give cosines and sines.

The real part 0 goes in the e, the 2 goes inside cos and sin.

α ± βi: e^(αx)(C₁cos βx + C₂sin βx)

So y = C₁cos(2x) + C₂sin(2x).

y = C1cos(2x) + C2sin(2x)

5 Check

Let's check one root, 2i.

Square it, with i² = −1: −4.

r = 2i: r² = −4

Into r² + 4: the numbers cancel and the i parts cancel.

0.

Right!

−4 + 4 = 0

Let's look back.

First: try y = e to the rx.

Then: the characteristic equation.

Then: solve for r.

Then: write the solution.

Last: check.

Answer
y = C1cos(2x) + C2sin(2x)

Your turn

Try this one yourself, then check:

Solve y″ + 9y = 0
Show the answer
y = C1cos(3x) + C2sin(3x)

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