Solve y'' - 3y' + 2y = 0, y(0) = 0, y'(0) = 1 using Laplace Step by Step

Step by step

The question

Question: solve y″ − 3y′ + 2y = 0, with y(0) = 0 and y′(0) = 1, using the Laplace transform.

y″ − 3y′ + 2y = 0

y(0) = 0, y′(0) = 1

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″} − 3ℒ{y′} + 2ℒ{y} = 0

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

(s²Y − 0 − 1) − 3(sY − 0) + 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 − 3s + 2).

Y(s² − 3s + 2) − 1 = 0

Divide by s² − 3s + 2, which factors into (s − 1)(s − 2).

Y = 1(s − 1)(s − 2)

3 Partial fractions

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

1(s − 1)(s − 2) = As − 1 + Bs − 2

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

Each letter keeps the factors it does not have.

1 = A(s − 2) + B(s − 1)

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

1 = −1 × A.

s = 1: 1 = −1 × A

So A = −1.

A = −1

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

1 = 1 × B.

s = 2: 1 = 1 × B

So B = 1.

B = 1

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

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

y = −e^(t) + e^(2t)

5 Check

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

Right!

y(0) = 0, y′(0) = 1

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

Right!

put in: −e^(t) + 4e^(2t) − 3(−e^(t) + 2e^(2t)) + 2(−e^(t) + e^(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