Solve y' + y = 3e^(2t), y(0) = 2 using Laplace Step by Step

Step by step

The question

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

y′ + y = 3e^(2t)

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)

Take the transform of both sides, term by term.

ℒ{y′} + ℒ{y} = ℒ{3e^(2t)}

Use the rules with y(0) = 2.

On the right, the table: 3/(s − 2).

(sY − 2) + Y = 3s − 2

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 + 1).

Y(s + 1) − 2 = 3s − 2

Move the starting part to the right.

Y(s + 1) = 3s − 2 + 2

One fraction on the right: times 2 by s − 2 over itself.

Y(s + 1) = 2s − 1(s − 2)

Divide by s + 1.

Y = 2s − 1(s + 1)(s − 2)

3 Partial fractions

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

2s − 1(s + 1)(s − 2) = As + 1 + Bs − 2

Times both sides by the whole bottom (s + 1)(s − 2).

Each letter keeps the factors it does not have.

2s − 1 = A(s − 2) + B(s + 1)

Put s = −1: every other letter is multiplied by 0.

−3 = −3 × A.

s = −1: −3 = −3 × A

So A = 1.

A = 1

Put s = 2: every other letter is multiplied by 0.

3 = 3 × B.

s = 2: 3 = 3 × B

So B = 1.

B = 1

So it splits into 1/(s + 1) + 1/(s − 2).

2s − 1(s + 1)(s − 2) = 1s + 1 + 1s − 2

4 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^(−t) + e^(2t).

y = e^(−t) + e^(2t)

5 Check

Let's check the starting values: y(0) = 2.

Right!

y(0) = 2

And put y into the equation: it gives 3e^(2t), the right side.

Right!

put in: −e^(−t) + 2e^(2t) + (e^(−t) + e^(2t)) = 3e^(2t)

Let's look back.

First: transform both sides.

Then: solve for Y.

Then: partial fractions.

Then: use the table backwards.

Last: check.

Answer