Solve dy/dx = 6cos(3x), y(0) = 1 Step by Step

Step by step

The question

Question: solve dy/dx = 6cos(3x), with y(0) = 1.

dydx = 6cos(3x)

y(0) = 1 is the starting value: when x is 0, y is 1.

y(0) = 1

1 The idea

dy/dx is only about x.

So y is a function whose derivative is 6cos(3x): an antiderivative.

y′ = 6cos(3x): y is a function whose derivative is 6cos(3x)

2 Integrate both sides

Integrate both sides.

The left side, the integral of dy/dx, is just y.

y = ∫ 6cos(3x) dx

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

∫ 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(3x)3 + C

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

y = 2sin(3x) + C

Careful!

Don't forget + C.

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

Careful: not y = 2sin(3x) ✗

3 Use the starting value

Put in x = 0 and y = 1.

1 = 2sin 0 + C

sin 0 = 0

Work out the numbers: 0.

1 = 0 + C

So C = 1.

C = 1

Put C back.

The answer: y = 2sin(3x) + 1.

y = 2sin(3x) + 1

4 Check

Let's check: the derivative of our answer is 6cos(3x).

The right side again.

Right!

y′ = 6cos(3x)

And at x = 0, y is 1: the starting value.

Right!

y(0) = 1

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer