Solve y'' + 4y = 0, y(0) = 3, y'(0) = -2 using Laplace Step by Step

Step by step

The question

Question: solve y″ + 4y = 0, with y(0) = 3 and y′(0) = −2, using the Laplace transform.

y″ + 4y = 0

y(0) = 3, y′(0) = −2

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) = 3 and y′(0) = −2.

(s²Y − 3s − (−2)) + 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) − (3s − 2) = 0

Divide by s² + 4.

Y = 3s − 2(s² + 4)

3 Split it up

Split the top: the s part and the number part.

Y = 3ss² + 4 − 2s² + 4

4 Use the table backwards

Each piece is already a row of the table.

cos(bt) → ss² + b²

sin(bt) → bs² + b²

Read the table backwards, piece by piece: 3cos(2t) − sin(2t).

y = 3cos(2t) − sin(2t)

5 Check

Let's check the starting values: y(0) = 3 and y′(0) = −2.

Right!

y(0) = 3, y′(0) = −2

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

Right!

put in: −12cos(2t) + 4sin(2t) + 4(3cos(2t) − sin(2t)) = 0

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: split it up.

Then: use the table backwards.

Last: check.

Answer