Solve y' + y = 3e^(2x), y(0) = 2 Step by Step

Step by step

The question

Question: solve y′ + y = 3e^(2x), with y(0) = 2.

y′ + y = 3e^(2x)

y(0) = 2 is the starting value.

y(0) = 2

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 = 3e^(2x).

P = 1, Q = 3e^(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 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 = 3e^(3x)

Careful!

The right side is multiplied by eˣ too.

Both sides, always.

Careful: not (e^xy)′ = 3e^(2x) ✗

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)′ = 3e^(3x)

5 Integrate both sides

Integrate both sides.

The integral of a derivative is the function itself.

e^xy = ∫ (3e^(3x)) dx

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

e^xy = e^(3x) + C

6 Solve for y

Divide both sides by eˣ.

y = e^(3x) + Ce^x

Divide each term.

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

y = e^(2x) + Ce^(−x)

C over e^x = Ce^(−x)

7 Use the starting value

Put in x = 0 and y = 2.

e⁰ = 1.

2 = 1 + C × 1

So C = 1.

C = 1

So y = e^(2x) + e^(−x).

y = e^(2x) + e^(−x)

8 Check

Let's check: put y back in.

First y′ is 2e^(2x) − e^(−x).

y′ = 2e^(2x) − e^(−x)

Then y′ + 1y.

y′ + y = 2e^(2x) − e^(−x) + 1(e^(2x) + e^(−x))

The e parts cancel, and we get 3e^(2x): Q again.

Right!

= 3e^(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.

Then: use the starting value.

Last: check.

Answer