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Solve y' + 2y = 0, y(0) = 3 using Laplace

Step by step, with the reason for every step · Solving with Laplace · Differential Equations

Step by step

The question

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

y′ + 2y = 0
y(0) = 3

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)

Take the transform of both sides, term by term.

ℒ{y′} + 2ℒ{y} = 0

Use the rules with y(0) = 3.

(sY − 3) + 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(s + 2).

Y(s + 2) − 3 = 0

Divide by s + 2.

Y = 3(s + 2)

3 Use the table backwards

Each piece is already a row of the table.

e^(at) → 1s − a

Read the table backwards, piece by piece: 3e^(−2t).

y = 3e^(−2t)

4 Check

Let's check the starting values: y(0) = 3.

Right!

y(0) = 3

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

Right!

put in: −6e^(−2t) + 2 × 3e^(−2t) = 0

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: use the table backwards.

Last: check.

Answer
y = 3e^(−2t)

Your turn

Try this one yourself, then check:

Solve y′ + 2y = 0
Show the answer
y = e^(−2t)

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