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Solve y'' + 10y' + 25y = 0

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

Step by step

The question

Question: solve y″ + 10y′ + 25y = 0.

y″ + 10y′ + 25y = 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) + 10re^(rx) + 25e^(rx) = 0

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

e^(rx)(r2 + 10r + 25) = 0

e to a power is never 0.

So the bracket must be 0.

e^(rx) is never 0

r² + 10r + 25 = 0: the characteristic equation.

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

r2 + 10r + 25 = 0

3 Solve for r

Two numbers that multiply to 25 and add to 10: 5 and 5, the same number twice.

5 × 5 = 25, 5 + 5 = 10

So it is a square: (r + 5)² = 0.

(r + 5)2 = 0

Only one root: r = −5, twice.

r = −5 (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^(−5x).

The second one needs an x.

Careful: not y = C1e^(−5x) + C2e^(−5x) ✗

So y = C₁e^(−5x) + C₂xe^(−5x).

y = C1e^(−5x) + C2xe^(−5x)

5 Check

Let's check the new part, xe^(−5x): its derivatives, by the product rule.

y = xe^(−5x): y′ = e^(−5x) − 5xe^(−5x), y″ = −10e^(−5x) + 25xe^(−5x)

Put them in: everything cancels, 0.

Right!

put in: −10e^(−5x) + 25xe^(−5x) + 10(e^(−5x) − 5xe^(−5x)) + 25 × xe^(−5x) = 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^(−5x) + C2xe^(−5x)

Your turn

Try this one yourself, then check:

Solve y″ − 6y′ + 9y = 0
Show the answer
y = C1e^(3x) + C2xe^(3x)

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