Solve y' - 3y = e^(2x) Step by Step | The integrating factor

Step by step

The question

Question: solve y′ − 3y = e^(2x).

y′ − 3y = e^(2x)

1 Standard form

Linear means this form: y′ + P y = Q.

P and Q have only x.

linear: y′ + P(x) y = Q(x)

Here P = −3 and Q = e^(2x).

P = −3, Q = e^(2x)

2 The integrating factor

The integrating factor is e to the integral of P.

Multiplying by it makes the left side one derivative.

integrating factor: μ = e^(∫P dx)

First the integral of P: the integral of −3 is −3x.

∫ −3 dx = −3x

So the integrating factor is e^(−3x).

No + C needed here: any one factor works.

μ = e^(−3x)

3 Multiply both sides

Multiply both sides by e^(−3x).

Every term, on both sides.

e^(−3x)y′ − 3e^(−3x)y = e^(−x)

Careful!

The right side is multiplied by e^(−3x) too.

Both sides, always.

Careful: not (e^(−3x)y)′ = e^(2x) ✗

4 One derivative on the left

By the product rule, the derivative of e^(−3x) × y is exactly our left side.

(μy)′ = μy′ + μ′y

μ′ = −−3e^(−3x)

So the left side is just (e^(−3x)y)′: one derivative.

(e^(−3x)y)′ = e^(−x)

5 Integrate both sides

Integrate both sides.

The integral of a derivative is the function itself.

e^(−3x)y = ∫ e^(−x) dx

On the right: −e^(−x), and + C.

e^(−3x)y = −e^(−x) + C

6 Solve for y

Divide both sides by e^(−3x).

y = −e^(−x) + Ce^(−3x)

Divide each term.

Dividing by e to a power is the same as e to minus that power.

y = −e^(2x) + Ce^(3x)

C over e^(−3x) = Ce^(3x)

7 Check

Let's check: put y back in.

First y′ is −2e^(2x) + 3Ce^(3x).

y′ = −2e^(2x) + 3Ce^(3x)

Then y′ − 3y.

y′ − 3y = −2e^(2x) + 3Ce^(3x) − 3(−e^(2x) + Ce^(3x))

The C parts cancel, and we get e^(2x): Q again.

Right!

= e^(2x)

Let's look back.

First: standard form.

Then: the integrating factor.

Then: multiply both sides.

Then: one derivative on the left.

Then: integrate both sides.

Then: solve for y.

Last: check.

Answer