Solve y' - 4y = 8, y(0) = -1 Step by Step

Step by step

The question

Question: solve y′ − 4y = 8, with y(0) = −1.

y′ − 4y = 8

y(0) = −1 is the starting value.

y(0) = −1

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 = −4 and Q = 8.

P = −4, Q = 8

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 −4 is −4x.

∫ −4 dx = −4x

So the integrating factor is e^(−4x).

No + C needed here: any one factor works.

μ = e^(−4x)

3 Multiply both sides

Multiply both sides by e^(−4x).

Every term, on both sides.

e^(−4x)y′ − 4e^(−4x)y = 8e^(−4x)

Careful!

The right side is multiplied by e^(−4x) too.

Both sides, always.

Careful: not (e^(−4x)y)′ = 8 ✗

4 One derivative on the left

By the product rule, the derivative of e^(−4x) × y is exactly our left side.

(μy)′ = μy′ + μ′y

μ′ = −−4e^(−4x)

So the left side is just (e^(−4x)y)′: one derivative.

(e^(−4x)y)′ = 8e^(−4x)

5 Integrate both sides

Integrate both sides.

The integral of a derivative is the function itself.

e^(−4x)y = ∫ (8e^(−4x)) dx

On the right: −2e^(−4x), and + C.

e^(−4x)y = −2e^(−4x) + C

6 Solve for y

Divide both sides by e^(−4x).

y = −2e^(−4x) + Ce^(−4x)

Divide each term.

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

y = −2 + Ce^(4x)

C over e^(−4x) = Ce^(4x)

7 Use the starting value

Put in x = 0 and y = −1.

e⁰ = 1.

−1 = −2 + C × 1

So C = 1.

C = 1

So y = −2 + e^(4x).

y = −2 + e^(4x)

8 Check

Let's check: put y back in.

First y′ is 4e^(4x).

y′ = 4e^(4x)

Then y′ − 4y.

y′ − 4y = 4e^(4x) − 4(−2 + e^(4x))

The e parts cancel, and we get 8: Q again.

Right!

= 8

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