Solve y'' + 4y = 8, y(0) = 0, y'(0) = 0 using Laplace Step by Step

Step by step

The question

Question: solve y″ + 4y = 8, with y(0) = 0 and y′(0) = 0, using the Laplace transform.

y″ + 4y = 8

y(0) = 0, 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″} + 4ℒ{y} = ℒ{8}

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

On the right, the table: 8 over s.

(s²Y − 0 − 0) + 4Y = 8s

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) = 8s

Divide by s² + 4.

Y = 8s(s² + 4)

3 Partial fractions

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

Over s² + b², the top can have an s.

8s(s² + 4) = As + Bs + Cs² + 4

Times both sides by the whole bottom s(s² + 4).

Each letter keeps the factors it does not have.

8 = A(s² + 4) + (Bs + C)s

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

8 = 4 × A.

s = 0: 8 = 4 × A

So A = 2.

A = 2

Now compare the s² parts on both sides.

s² parts: 0 = A + B

So B = −2.

B = −2

And the s parts.

s parts: 0 = C

So C = 0.

C = 0

So it splits into 2/s − 2s/(s² + 4).

8s(s² + 4) = 2s − 2ss² + 4

4 Use the table backwards

Each piece is already a row of the table.

1 → 1s

cos(bt) → ss² + b²

Read the table backwards, piece by piece: 2 − 2cos(2t).

y = 2 − 2cos(2t)

5 Check

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

Right!

y(0) = 0, y′(0) = 0

And put y into the equation: it gives 8, the right side.

Right!

put in: 8cos(2t) + 4(2 − 2cos(2t)) = 8

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer