Solve y' = 12x^2 - 6x + 1, y(1) = 0 Step by Step

Step by step

The question

Question: solve y′ = 12x² − 6x + 1, with y(1) = 0.

y′ = 12x² − 6x + 1

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

y(1) = 0

1 The idea

y′ is only about x.

So y is a function whose derivative is 12x² − 6x + 1: an antiderivative.

y′ = 12x² − 6x + 1: y is a function whose derivative is 12x² − 6x + 1

2 Integrate both sides

Integrate both sides.

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

y = ∫ (12x² − 6x + 1) dx

The power rule backwards: add 1 to the power, divide by the new power.

∫ xⁿ dx = x^{n + 1}n + 1

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

y = 12 × x³3 − 6 × x²2 + x1 + C

Tidy the numbers: y = 4x³ − 3x² + x + C.

y = 4x³ − 3x² + x + C

Careful!

Don't forget + C.

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

Careful: not y = 4x³ − 3x² + x ✗

3 Use the starting value

Put in x = 1 and y = 0.

0 = 4(1)³ − 3(1)² + (1) + C

Work out the numbers: 2.

0 = 2 + C

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

0 − 2 = C

So C = −2.

C = −2

Put C back.

The answer: y = 4x³ − 3x² + x − 2.

y = 4x³ − 3x² + x − 2

4 Check

Let's check: the derivative of our answer is 12x² − 6x + 1.

The right side again.

Right!

y′ = 12x² − 6x + 1

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

Right!

y(1) = 0

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer