Steply

Solve y'' + y' - 6y = 0, y(0) = 1, y'(0) = -8

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

Step by step

The question

Question: solve y″ + y′ − 6y = 0, with y(0) = 1 and y′(0) = −8.

y″ + y′ − 6y = 0

The starting values: at x = 0, y is 1 and its slope y′ is −8.

y(0) = 1, y′(0) = −8
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) + re^(rx) − 6e^(rx) = 0

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

e^(rx)(r2 + r − 6) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

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

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

r2 + r − 6 = 0

3 Solve for r

Two numbers that multiply to −6 and add to 1: 3 and −2.

3 × (−2) = −6, 3 + (−2) = 1

So it factors: (r + 3)(r − 2) = 0.

(r + 3)(r − 2) = 0

Each bracket can be 0: r = −3 or r = 2.

r = −3 or r = 2

Careful!

(r + 3) is 0 when r is −3: the opposite sign of the number in the bracket.

Careful: not r = 3, −2 ✗

4 Write the solution

Each root gives a solution e^(rx).

Any mix of the two works, so we add them with two numbers, C₁ and C₂.

two roots r₁, r₂: C₁e^(r₁x) + C₂e^(r₂x)

So y = C₁e^(−3x) + C₂e^(2x).

y = C1e^(−3x) + C2e^(2x)

5 Use the starting values

We need y′ too: the derivative of our y.

y′ = −3C1e^(−3x) + 2C2e^(2x)
at x = 0: e^(0) = 1,

Put x = 0 into y: C₁ + C₂ = 1.

① C1 + C2 = 1

Put x = 0 into y′: −3C₁ + 2C₂ = −8.

② −3C1 + 2C2 = −8

From ①: C₁ = 1 − C₂.

C1 = 1 − C2

Put it into ②.

−3(1 − C2) + 2C2 = −8

Open the bracket and collect: −3 × 1 moves over.

5C2 = −5

So C₂ = −1, and C₁ = 2.

C2 = −1

Put them in: y = 2e^(−3x) − e^(2x).

y = 2e^(−3x) − e^(2x)

6 Check

Let's check: each root must make r² + r − 6 = 0.

r = −3 gives 0.

r = −3: (−3)² + (−3) − 6 = 0

And r = 2 gives 0.

Right!

r = 2: 2² + 2 − 6 = 0

And the starting values: y(0) = 1 and y′(0) = −8.

Right!

y(0) = 1, y′(0) = −8

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.

Answer
y = 2e^(−3x) − e^(2x)

Your turn

Try this one yourself, then check:

Solve y″ − 6y′ + 8y = 0
Show the answer
y = −2e^(2x) + 2e^(4x)

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