Solve y'' - 4y = 0, y(0) = 1, y'(0) = 6 using Laplace Step by Step

Step by step

The question

Question: solve y″ − 4y = 0, with y(0) = 1 and y′(0) = 6, using the Laplace transform.

y″ − 4y = 0

y(0) = 1, y′(0) = 6

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″} − 4ℒ{y} = 0

Use the rules with y(0) = 1 and y′(0) = 6.

(s²Y − 1s − 6) − 4Y = 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 − 4).

Y(s² − 4) − (s + 6) = 0

Divide by s² − 4, which factors into (s + 2)(s − 2).

Y = s + 6(s + 2)(s − 2)

3 Partial fractions

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

s + 6(s + 2)(s − 2) = As + 2 + Bs − 2

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

Each letter keeps the factors it does not have.

s + 6 = A(s − 2) + B(s + 2)

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

4 = −4 × A.

s = −2: 4 = −4 × A

So A = −1.

A = −1

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

8 = 4 × B.

s = 2: 8 = 4 × B

So B = 2.

B = 2

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

s + 6(s + 2)(s − 2) = −1s + 2 + 2s − 2

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^(2t).

y = −e^(−2t) + 2e^(2t)

5 Check

Let's check the starting values: y(0) = 1 and y′(0) = 6.

Right!

y(0) = 1, y′(0) = 6

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

Right!

put in: −4e^(−2t) + 8e^(2t) − 4(−e^(−2t) + 2e^(2t)) = 0

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer