Solve y' = 9x^2 - 4x + 3, y(-1) = 1 Step by Step

Step by step

The question

Question: solve y′ = 9x² − 4x + 3, with y(−1) = 1.

y′ = 9x² − 4x + 3

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

y(−1) = 1

1 The idea

y′ is only about x.

So y is a function whose derivative is 9x² − 4x + 3: an antiderivative.

y′ = 9x² − 4x + 3: y is a function whose derivative is 9x² − 4x + 3

2 Integrate both sides

Integrate both sides.

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

y = ∫ (9x² − 4x + 3) 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 = 9 × x³3 − 4 × x²2 + 3 × x1 + C

Tidy the numbers: y = 3x³ − 2x² + 3x + C.

y = 3x³ − 2x² + 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 = 3x³ − 2x² + 3x ✗

3 Use the starting value

Put in x = −1 and y = 1.

1 = 3(−1)³ − 2(−1)² + 3(−1) + C

Work out the numbers: −8.

1 = −8 + C

To get C alone, add 8 on both sides.

1 + 8 = C

So C = 9.

C = 9

Put C back.

The answer: y = 3x³ − 2x² + 3x + 9.

y = 3x³ − 2x² + 3x + 9

4 Check

Let's check: the derivative of our answer is 9x² − 4x + 3.

The right side again.

Right!

y′ = 9x² − 4x + 3

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

Right!

y(−1) = 1

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer