Solve y' - y = 2e^(3t), y(0) = 1 using Laplace Step by Step

Step by step

The question

Question: solve y′ − y = 2e^(3t), with y(0) = 1, using the Laplace transform.

y′ − y = 2e^(3t)

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′} − ℒ{y} = ℒ{2e^(3t)}

Use the rules with y(0) = 1.

On the right, the table: 2/(s − 3).

(sY − 1) − Y = 2s − 3

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 − 1).

Y(s − 1) − 1 = 2s − 3

Move the starting part to the right.

Y(s − 1) = 2s − 3 + 1

One fraction on the right: times 1 by s − 3 over itself.

Y(s − 1) = s − 1(s − 3)

Divide by s − 1.

Y = s − 1(s − 1)(s − 3)

3 Partial fractions

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

s − 1(s − 1)(s − 3) = As − 1 + Bs − 3

Times both sides by the whole bottom (s − 1)(s − 3).

Each letter keeps the factors it does not have.

s − 1 = A(s − 3) + B(s − 1)

Put s = 1: every other letter is multiplied by 0.

0 = −2 × A.

s = 1: 0 = −2 × A

So A = 0.

A = 0

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

2 = 2 × B.

s = 3: 2 = 2 × B

So B = 1.

B = 1

So it splits into 1/(s − 3).

s − 1(s − 1)(s − 3) = 1s − 3

4 Use the table backwards

Each piece is already a row of the table.

e^(at) → 1s − a

Read the table backwards, piece by piece: e^(3t).

y = 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 2e^(3t), the right side.

Right!

put in: 3e^(3t) − e^(3t) = 2e^(3t)

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer