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

Step by step

The question

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

y″ − y = 0

y(0) = 2, 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} = 0

Use the rules with y(0) = 2 and y′(0) = 0.

(s²Y − 2s − 0) − Y = 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 − 1).

Y(s² − 1) − 2s = 0

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

Y = 2s(s + 1)(s − 1)

3 Partial fractions

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

2s(s + 1)(s − 1) = As + 1 + Bs − 1

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

Each letter keeps the factors it does not have.

2s = A(s − 1) + B(s + 1)

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

−2 = −2 × A.

s = −1: −2 = −2 × A

So A = 1.

A = 1

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

2 = 2 × B.

s = 1: 2 = 2 × B

So B = 1.

B = 1

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

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

y = e^(−t) + e^(t)

5 Check

Let's check the starting values: y(0) = 2 and y′(0) = 0.

Right!

y(0) = 2, y′(0) = 0

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

Right!

put in: e^(−t) + e^(t) − (e^(−t) + e^(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