Solve y' = 6cos(2x) - 4x, y(0) = 2 Step by Step

Step by step

The question

Question: solve y′ = 6cos(2x) − 4x, with y(0) = 2.

y′ = 6cos(2x) − 4x

y(0) = 2 is the starting value: when x is 0, y is 2.

y(0) = 2

1 The idea

y′ is only about x.

So y is a function whose derivative is 6cos(2x) − 4x: an antiderivative.

y′ = 6cos(2x) − 4x: y is a function whose derivative is 6cos(2x) − 4x

2 Integrate both sides

Integrate both sides.

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

y = ∫ (6cos(2x) − 4x) dx

The rules, one for each kind of term.

∫ xⁿ dx = x^{n + 1}n + 1

∫ cos(bx) dx = sin(bx)b

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

y = 6 × sin(2x)2 − 4 × x²2 + C

Tidy the numbers: y = 3sin(2x) − 2x² + C.

y = 3sin(2x) − 2x² + C

Careful!

Don't forget + C.

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

Careful: not y = 3sin(2x) − 2x² ✗

3 Use the starting value

Put in x = 0 and y = 2.

2 = 3sin 0 − 2(0)² + C

sin 0 = 0

Work out the numbers: 0.

2 = 0 + C

So C = 2.

C = 2

Put C back.

The answer: y = 3sin(2x) − 2x² + 2.

y = 3sin(2x) − 2x² + 2

4 Check

Let's check: the derivative of our answer is 6cos(2x) − 4x.

The right side again.

Right!

y′ = 6cos(2x) − 4x

And at x = 0, y is 2: the starting value.

Right!

y(0) = 2

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer