Step by step
The question
Question: solve y′ + 4y = 0, with y(0) = 1, using the Laplace transform.
y′ + 4y = 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)
Take the transform of both sides, term by term.
ℒ{y′} + 4ℒ{y} = 0
Use the rules with y(0) = 1.
(sY − 1) + 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(s + 4).
Y(s + 4) − 1 = 0
Divide by s + 4.
Y = 1(s + 4)
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: e^(−4t).
y = e^(−4t)
4 Check
Let's check the starting values: y(0) = 1.
Right!
y(0) = 1
And put y into the equation: it gives 0, the right side.
Right!
put in: −4e^(−4t) + 4 × e^(−4t) = 0
Let's look back.
First: transform both sides.
Then: solve for Y.
Then: use the table backwards.
Last: check.
Answer
y = e^(−4t)