Solve y' - 2y = 6e^(-t), y(0) = 0 using Laplace Step by Step

Step by step

The question

Question: solve y′ − 2y = 6e^(−t), with y(0) = 0, using the Laplace transform.

y′ − 2y = 6e^(−t)

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{6e^(−t)}

Use the rules with y(0) = 0.

On the right, the table: 6/(s + 1).

(sY − 0) − 2Y = 6s + 1

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) = 6s + 1

Divide by s − 2.

Y = 6(s − 2)(s + 1)

3 Partial fractions

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

6(s − 2)(s + 1) = As − 2 + Bs + 1

Times both sides by the whole bottom (s − 2)(s + 1).

Each letter keeps the factors it does not have.

6 = A(s + 1) + B(s − 2)

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

6 = 3 × A.

s = 2: 6 = 3 × A

So A = 2.

A = 2

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

6 = −3 × B.

s = −1: 6 = −3 × B

So B = −2.

B = −2

So it splits into 2/(s − 2) − 2/(s + 1).

6(s − 2)(s + 1) = 2s − 2 − 2s + 1

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: 2e^(2t) − 2e^(−t).

y = 2e^(2t) − 2e^(−t)

5 Check

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

Right!

y(0) = 0

And put y into the equation: it gives 6e^(−t), the right side.

Right!

put in: 4e^(2t) + 2e^(−t) − 2(2e^(2t) − 2e^(−t)) = 6e^(−t)

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer