Steply

Solve y'' - 4y = 0

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

Step by step

The question

Question: solve y″ − 4y = 0.

y″ − 4y = 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) − 4e^(rx) = 0

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

e^(rx)(r2 − 4) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

r² − 4 = 0: the characteristic equation.

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

r2 − 4 = 0

3 Solve for r

Two numbers that multiply to −4 and add to 0: 2 and −2.

2 × (−2) = −4, 2 + (−2) = 0

So it factors: (r + 2)(r − 2) = 0.

(r + 2)(r − 2) = 0

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

r = −2 or r = 2

Careful!

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

Careful: not r = 2, −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^(−2x) + C₂e^(2x).

y = C1e^(−2x) + C2e^(2x)

5 Check

Let's check: each root must make r² − 4 = 0.

r = −2 gives 0.

r = −2: (−2)² − 4 = 0

And r = 2 gives 0.

Right!

r = 2: 2² − 4 = 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^(2x)

Your turn

Try this one yourself, then check:

Solve y″ + 2y′ − 3y = 0
Show the answer
y = C1e^(−3x) + C2ex

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