Steply

Solve y'' - 7y' + 12y = 0

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

Step by step

The question

Question: solve y″ − 7y′ + 12y = 0.

y″ − 7y′ + 12y = 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) − 7re^(rx) + 12e^(rx) = 0

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

e^(rx)(r2 − 7r + 12) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

r² − 7r + 12 = 0: the characteristic equation.

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

r2 − 7r + 12 = 0

3 Solve for r

Two numbers that multiply to 12 and add to −7: −3 and −4.

−3 × (−4) = 12, −3 + (−4) = −7

So it factors: (r − 3)(r − 4) = 0.

(r − 3)(r − 4) = 0

Each bracket can be 0: r = 3 or r = 4.

r = 3 or r = 4

Careful!

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

Careful: not r = −3, −4 ✗

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^(4x).

y = C1e^(3x) + C2e^(4x)

5 Check

Let's check: each root must make r² − 7r + 12 = 0.

r = 3 gives 0.

r = 3: 3² − 7 × 3 + 12 = 0

And r = 4 gives 0.

Right!

r = 4: 4² − 7 × 4 + 12 = 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^(3x) + C2e^(4x)

Your turn

Try this one yourself, then check:

Solve y″ − 5y′ + 6y = 0
Show the answer
y = C1e^(2x) + C2e^(3x)

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