Solve y'' + 6y' + 10y = 0 Step by Step

Step by step

The question

Question: solve y″ + 6y′ + 10y = 0.

y″ + 6y′ + 10y = 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) + 6re^(rx) + 10e^(rx) = 0

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

e^(rx)(r² + 6r + 10) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

r² + 6r + 10 = 0: the characteristic equation.

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

r² + 6r + 10 = 0

3 Solve for r

It does not factor nicely, so we use the formula, with a = 1, b = 6, c = 10.

r = −b ± √b² − 4ac2a

Put the numbers in.

r = −6 ± √6² − 4 × 1 × 102 × 1

Under the root: −4.

r = −6 ± √−42

A minus under the root: √(−4) = 2i, because i = √(−1).

√−4 = √4 × √−1 = 2i

r = −6 ± 2i2

Divide both parts by 2: r = −3 ± i.

Two complex roots.

r = −3 ± i

Careful!

Don't lose the real part, −3.

It gives e^(−3x) in front.

Careful: not y = C₁cos(x) + C₂sin(x) ✗

4 Write the solution

Complex roots give cosines and sines.

The real part −3 goes in the e, the 1 goes inside cos and sin.

α ± βi: e^(αx)(C₁cos βx + C₂sin βx)

So y = e^(−3x)(C₁cos x + C₂sin x).

y = e^(−3x)(C₁cos x + C₂sin x)

5 Check

Let's check one root, −3 + i.

Square it, with i² = −1: 8 − 6i.

r = −3 + i: r² = 8 − 6i

Into r² + 6r + 10: the numbers cancel and the i parts cancel.

Right!

8 − 6i + 6(−3 + i) + 10 = 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