Steply

Solve 9y'' - 6y' + y = 0

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

Step by step

The question

Question: solve 9y″ − 6y′ + y = 0.

9y″ − 6y′ + y = 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.

9r2e^(rx) − 6re^(rx) + e^(rx) = 0

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

e^(rx)(9r2 − 6r + 1) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

9r² − 6r + 1 = 0: the characteristic equation.

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

9r2 − 6r + 1 = 0

3 Solve for r

It does not factor nicely, so we use the formula, with a = 9, b = −6, c = 1.

r = −b ± √b² − 4ac2a

Put the numbers in.

r = 6 ± √(−6)² − 4 × 9 × 12 × 9

Under the root: 0.

r = 6 ± √018

Under the root is 0, so plus and minus give the same: r = 1/3, twice.

r = 1/3 (twice)

4 Write the solution

One root gives only one solution.

The second one is the same times x.

(Put it in: it works.)

one root twice: C₁e^(rx) + C₂xe^(rx)

Careful!

That is only one solution twice: (C₁ + C₂)e^(x/3).

The second one needs an x.

Careful: not y = C1e^(x/3) + C2e^(x/3) ✗

So y = C₁e^(x/3) + C₂xe^(x/3).

y = C1e^(x/3) + C2xe^(x/3)

5 Check

Let's check the new part, xe^(x/3): its derivatives, by the product rule.

y = xe^(x/3): y′ = e^(x/3) + 13xe^(x/3), y″ = 23e^(x/3) + 19xe^(x/3)

Put them in: everything cancels, 0.

Right!

put in: 9(23e^(x/3) + 19xe^(x/3)) − 6(e^(x/3) + 13xe^(x/3)) + xe^(x/3) = 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 = C1e^(x/3) + C2xe^(x/3)

Your turn

Try this one yourself, then check:

Solve 9y″ − 12y′ + 4y = 0
Show the answer
y = C1e^(2x/3) + C2xe^(2x/3)

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