Solve y' + 3y = 6, y(0) = 1 using Laplace Step by Step

Step by step

The question

Question: solve y′ + 3y = 6, with y(0) = 1, using the Laplace transform.

y′ + 3y = 6

y(0) = 1

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′} + 3ℒ{y} = ℒ{6}

Use the rules with y(0) = 1.

On the right, the table: 6 over s.

(sY − 1) + 3Y = 6s

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 + 3).

Y(s + 3) − 1 = 6s

Move the starting part to the right.

Y(s + 3) = 6s + 1

One fraction on the right: times 1 by s over itself.

Y(s + 3) = s + 6s

Divide by s + 3.

Y = s + 6s(s + 3)

3 Partial fractions

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

s + 6s(s + 3) = As + Bs + 3

Times both sides by the whole bottom s(s + 3).

Each letter keeps the factors it does not have.

s + 6 = A(s + 3) + Bs

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

6 = 3 × A.

s = 0: 6 = 3 × A

So A = 2.

A = 2

Put s = −3: every other letter is multiplied by 0.

3 = −3 × B.

s = −3: 3 = −3 × B

So B = −1.

B = −1

So it splits into 2/s − 1/(s + 3).

s + 6s(s + 3) = 2s − 1s + 3

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 − e^(−3t).

y = 2 − e^(−3t)

5 Check

Let's check the starting values: y(0) = 1.

Right!

y(0) = 1

And put y into the equation: it gives 6, the right side.

Right!

put in: 3e^(−3t) + 3(2 − e^(−3t)) = 6

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer