Worked examples
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.
▶ Watch it solved
The question
Question: solve y′ + 3y = 6, with y(0) = 1, using the Laplace transform.
y′ + 3y = 6
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′} + 3ℒ{y} = ℒ{6}
Use the rules with y(0) = 1.
On the right, the table: 6 over s.
(sY − 1) + 3Y = 6s
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 + 3).
Y(s + 3) − 1 = 6s
Move the starting part to the right.
Y(s + 3) = 6s + 1
One fraction on the right: times 1 by s over itself.
Y(s + 3) = s + 6s
Divide by s + 3.
Y = s + 6s(s + 3)
3 Partial fractions
Split it into simple fractions, one for each factor at the bottom.
s + 6s(s + 3) = As + Bs + 3
Times both sides by the whole bottom s(s + 3).
Each letter keeps the factors it does not have.
s + 6 = A(s + 3) + Bs
Put s = 0: every other letter is multiplied by 0.
6 = 3 × A.
s = 0: 6 = 3 × A
So A = 2.
A = 2
Put s = −3: every other letter is multiplied by 0.
3 = −3 × B.
s = −3: 3 = −3 × B
So B = −1.
B = −1
So it splits into 2/s − 1/(s + 3).
s + 6s(s + 3) = 2s − 1s + 3
4 Use the table backwards
Each piece is already a row of the table.
1 → 1s
e^(at) → 1s − a
Read the table backwards, piece by piece: 2 − e^(−3t).
y = 2 − e^(−3t)
5 Check
Let's check the starting values: y(0) = 1.
Right!
y(0) = 1
And put y into the equation: it gives 6, the right side.
Right!
put in: 3e^(−3t) + 3(2 − e^(−3t)) = 6
Let's look back.
First: transform both sides.
Then: solve for Y.
Then: partial fractions.
Then: use the table backwards.
Last: check.
▶ Watch it solved
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.
▶ Watch it solved
