Solve y' + 2y = 4, y(0) = 0 using Laplace Step by Step

Step by step

The question

Question: solve y′ + 2y = 4, with y(0) = 0, using the Laplace transform.

y′ + 2y = 4

y(0) = 0

1 Transform both sides

Call the transform of y, Y.

The derivative rules bring in the starting values: that is why Laplace is so handy here.

ℒ{y} = Y

ℒ{y′} = sY − y(0)

Take the transform of both sides, term by term.

ℒ{y′} + 2ℒ{y} = ℒ{4}

Use the rules with y(0) = 0.

On the right, the table: 4 over s.

(sY − 0) + 2Y = 4s

Careful!

Don't forget the starting value: ℒ{y′} is sY minus y(0).

Careful: not ℒ{y′} = sY ✗

2 Solve for Y

Put the Y terms together: Y(s + 2).

Y(s + 2) = 4s

Divide by s + 2.

Y = 4s(s + 2)

3 Partial fractions

Split it into simple fractions, one for each factor at the bottom.

4s(s + 2) = As + Bs + 2

Times both sides by the whole bottom s(s + 2).

Each letter keeps the factors it does not have.

4 = A(s + 2) + Bs

Put s = 0: every other letter is multiplied by 0.

4 = 2 × A.

s = 0: 4 = 2 × A

So A = 2.

A = 2

Put s = −2: every other letter is multiplied by 0.

4 = −2 × B.

s = −2: 4 = −2 × B

So B = −2.

B = −2

So it splits into 2/s − 2/(s + 2).

4s(s + 2) = 2s − 2s + 2

4 Use the table backwards

Each piece is already a row of the table.

1 → 1s

e^(at) → 1s − a

Read the table backwards, piece by piece: 2 − 2e^(−2t).

y = 2 − 2e^(−2t)

5 Check

Let's check the starting values: y(0) = 0.

Right!

y(0) = 0

And put y into the equation: it gives 4, the right side.

Right!

put in: 4e^(−2t) + 2(2 − 2e^(−2t)) = 4

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer