Solve y'' + y' - 2y = 0, y(0) = 3, y'(0) = 0 using Laplace Step by Step

Step by step

The question

Question: solve y″ + y′ − 2y = 0, with y(0) = 3 and y′(0) = 0, using the Laplace transform.

y″ + y′ − 2y = 0

y(0) = 3, 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)

ℒ{y″} = s²Y − sy(0) − y′(0)

Take the transform of both sides, term by term.

ℒ{y″} + ℒ{y′} − 2ℒ{y} = 0

Use the rules with y(0) = 3 and y′(0) = 0.

(s²Y − 3s − 0) + (sY − 3) − 2Y = 0

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(s² + s − 2).

Y(s² + s − 2) − (3s + 3) = 0

Divide by s² + s − 2, which factors into (s + 2)(s − 1).

Y = 3s + 3(s + 2)(s − 1)

3 Partial fractions

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

3s + 3(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.

3s + 3 = A(s − 1) + B(s + 2)

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

−3 = −3 × A.

s = −2: −3 = −3 × A

So A = 1.

A = 1

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 1/(s + 2) + 2/(s − 1).

3s + 3(s + 2)(s − 1) = 1s + 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: e^(−2t) + 2e^(t).

y = e^(−2t) + 2e^(t)

5 Check

Let's check the starting values: y(0) = 3 and y′(0) = 0.

Right!

y(0) = 3, y′(0) = 0

And put y into the equation: it gives 0, the right side.

Right!

put in: 4e^(−2t) + 2e^(t) + (−2e^(−2t) + 2e^(t)) − 2(e^(−2t) + 2e^(t)) = 0

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer