Solve y'' - 5y' + 6y = 0, y(0) = 1, y'(0) = 4 using Laplace Step by Step

Step by step

The question

Question: solve y″ − 5y′ + 6y = 0, with y(0) = 1 and y′(0) = 4, using the Laplace transform.

y″ − 5y′ + 6y = 0

y(0) = 1, y′(0) = 4

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″} − 5ℒ{y′} + 6ℒ{y} = 0

Use the rules with y(0) = 1 and y′(0) = 4.

(s²Y − 1s − 4) − 5(sY − 1) + 6Y = 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 − 5s + 6).

Y(s² − 5s + 6) − (s − 1) = 0

Divide by s² − 5s + 6, which factors into (s − 2)(s − 3).

Y = s − 1(s − 2)(s − 3)

3 Partial fractions

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

s − 1(s − 2)(s − 3) = As − 2 + Bs − 3

Times both sides by the whole bottom (s − 2)(s − 3).

Each letter keeps the factors it does not have.

s − 1 = A(s − 3) + B(s − 2)

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

1 = −1 × A.

s = 2: 1 = −1 × A

So A = −1.

A = −1

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

2 = 1 × B.

s = 3: 2 = 1 × B

So B = 2.

B = 2

So it splits into −1/(s − 2) + 2/(s − 3).

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

y = −e^(2t) + 2e^(3t)

5 Check

Let's check the starting values: y(0) = 1 and y′(0) = 4.

Right!

y(0) = 1, y′(0) = 4

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

Right!

put in: −4e^(2t) + 18e^(3t) − 5(−2e^(2t) + 6e^(3t)) + 6(−e^(2t) + 2e^(3t)) = 0

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer