Solve y' + y = e^x Step by Step | The integrating factor

Step by step

The question

Question: solve y′ + y = eˣ.

y′ + y = e^x

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 = 1 and Q = eˣ.

P = 1, Q = e^x

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 1 is x.

∫ 1 dx = x

So the integrating factor is eˣ.

No + C needed here: any one factor works.

μ = e^x

3 Multiply both sides

Multiply both sides by eˣ.

Every term, on both sides.

e^xy′ + e^xy = e^(2x)

Careful!

The right side is multiplied by eˣ too.

Both sides, always.

Careful: not (e^xy)′ = e^x ✗

4 One derivative on the left

By the product rule, the derivative of eˣ × y is exactly our left side.

(μy)′ = μy′ + μ′y

μ′ = e^x

So the left side is just (eˣy)′: one derivative.

(e^xy)′ = e^(2x)

5 Integrate both sides

Integrate both sides.

The integral of a derivative is the function itself.

e^xy = ∫ e^(2x) dx

On the right: ½e^(2x), and + C.

e^xy = 12e^(2x) + C

6 Solve for y

Divide both sides by eˣ.

y = 12e^(2x) + Ce^x

Divide each term.

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

y = 12e^x + Ce^(−x)

C over e^x = Ce^(−x)

7 Check

Let's check: put y back in.

First y′ is ½eˣ − Ce^(−x).

y′ = 12e^x − Ce^(−x)

Then y′ + 1y.

y′ + y = 12e^x − Ce^(−x) + 1(12e^x + Ce^(−x))

The C parts cancel, and we get eˣ: Q again.

Right!

= e^x

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