Solve y' = 4sin(2x), y(0) = 0 Step by Step

Step by step

The question

Question: solve y′ = 4sin(2x), with y(0) = 0.

y′ = 4sin(2x)

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

y(0) = 0

1 The idea

y′ is only about x.

So y is a function whose derivative is 4sin(2x): an antiderivative.

y′ = 4sin(2x): y is a function whose derivative is 4sin(2x)

2 Integrate both sides

Integrate both sides.

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

y = ∫ 4sin(2x) dx

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

∫ sin(bx) dx = −cos(bx)b

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

y = 4 × (−cos(2x)2) + C

Tidy the numbers: y = −2cos(2x) + C.

y = −2cos(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 = −2cos(2x) ✗

3 Use the starting value

Put in x = 0 and y = 0.

0 = −2cos 0 + C

cos 0 = 1

Work out the numbers: −2.

0 = −2 + C

To get C alone, add 2 on both sides.

0 + 2 = C

So C = 2.

C = 2

Put C back.

The answer: y = −2cos(2x) + 2.

y = −2cos(2x) + 2

4 Check

Let's check: the derivative of our answer is 4sin(2x).

The right side again.

Right!

y′ = 4sin(2x)

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

Right!

y(0) = 0

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer