Solve y' = e^(2x), y(0) = 3 Step by Step

Step by step

The question

Question: solve y′ = e^(2x), with y(0) = 3.

y′ = e^(2x)

y(0) = 3 is the starting value: when x is 0, y is 3.

y(0) = 3

1 The idea

y′ is only about x.

So y is a function whose derivative is e^(2x): an antiderivative.

y′ = e^(2x): y is a function whose derivative is e^(2x)

2 Integrate both sides

Integrate both sides.

The left side, the integral of y′, is just y.

y = ∫ e^(2x) dx

The rule: divide by the number in front of x.

∫ e^kx dx = e^kxk

Each term by its rule, and one + C: the number we cannot see after a derivative.

y = e^(2x)2 + C

Tidy the numbers: y = ½e^(2x) + C.

y = 12e^(2x) + C

Careful!

Don't forget + C.

Every number C gives another solution, and we need C to use the starting value.

Careful: not y = 12e^(2x) ✗

3 Use the starting value

Put in x = 0 and y = 3.

3 = 12e^(0) + C

e^(0) = 1

Work out the numbers: 1/2.

3 = 12 + C

To get C alone, take away 1/2 on both sides.

3 − 12 = C

So C = 5/2.

C = 52

Put C back.

The answer: y = ½e^(2x) + 5/2.

y = 12e^(2x) + 52

4 Check

Let's check: the derivative of our answer is e^(2x).

The right side again.

Right!

y′ = e^(2x)

And at x = 0, y is 3: the starting value.

Right!

y(0) = 3

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer