Solve y' + 2y = 6e^x, y(0) = 5

Step by step

The question

Question: solve y′ + 2y = 6eˣ, with y(0) = 5.

y(0) = 5 is the starting value.

1 Standard form

Linear means this form: y′ + P y = Q.

P and Q have only x.

Here P = 2 and Q = 6eˣ.

2 The integrating factor

The integrating factor is e to the integral of P.

Multiplying by it makes the left side one derivative.

First the integral of P: the integral of 2 is 2x.

So the integrating factor is e^(2x).

No + C needed here: any one factor works.

3 Multiply both sides

Multiply both sides by e^(2x).

Every term, on both sides.

Careful!

The right side is multiplied by e^(2x) too.

Both sides, always.

4 One derivative on the left

By the product rule, the derivative of e^(2x) × y is exactly our left side.

So the left side is just (e^(2x)y)′: one derivative.

5 Integrate both sides

Integrate both sides.

The integral of a derivative is the function itself.

On the right: 2e^(3x), and + C.

6 Solve for y

Divide both sides by e^(2x).

Divide each term.

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

7 Use the starting value

Put in x = 0 and y = 5.

e⁰ = 1.

So C = 3.

So y = 2eˣ + 3e^(−2x).

8 Check

Let's check: put y back in.

First y′ is 2eˣ − 6e^(−2x).

Then y′ + 2y.

The e parts cancel, and we get 6eˣ: Q again.

Right!

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.

Then: use the starting value.

Last: check.