Steply

Solve y'' - 5y' + 6y = 0

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

Step by step

The question

Question: solve y″ − 5y′ + 6y = 0.

y″ − 5y′ + 6y = 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) − 5re^(rx) + 6e^(rx) = 0

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

e^(rx)(r2 − 5r + 6) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

r² − 5r + 6 = 0: the characteristic equation.

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

r2 − 5r + 6 = 0

3 Solve for r

Two numbers that multiply to 6 and add to −5: −2 and −3.

−2 × (−3) = 6, −2 + (−3) = −5

So it factors: (r − 2)(r − 3) = 0.

(r − 2)(r − 3) = 0

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

r = 2 or r = 3

Careful!

(r − 2) is 0 when r is 2: the opposite sign of the number in the bracket.

Careful: not r = −2, −3 ✗

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^(2x) + C₂e^(3x).

y = C1e^(2x) + C2e^(3x)

5 Check

Let's check: each root must make r² − 5r + 6 = 0.

r = 2 gives 0.

r = 2: 2² − 5 × 2 + 6 = 0

And r = 3 gives 0.

Right!

r = 3: 3² − 5 × 3 + 6 = 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^(2x) + C2e^(3x)

Your turn

Try this one yourself, then check:

Solve y″ − 5y′ + 4y = 0
Show the answer
y = C1ex + C2e^(4x)

More from Second order

The whole lesson: Second order →