Solve y' = 2x + 1, y(0) = 4 Step by Step

Step by step

The question

Question: solve y′ = 2x + 1, with y(0) = 4.

y′ = 2x + 1

y(0) = 4 is the starting value: when x is 0, y is 4.

y(0) = 4

1 The idea

y′ is only about x.

So y is a function whose derivative is 2x + 1: an antiderivative.

y′ = 2x + 1: y is a function whose derivative is 2x + 1

2 Integrate both sides

Integrate both sides.

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

y = ∫ (2x + 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 = 2 × x²2 + x1 + C

Tidy the numbers: y = x² + x + C.

y = x² + 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 = x² + x ✗

3 Use the starting value

Put in x = 0 and y = 4.

4 = (0)² + (0) + C

Work out the numbers: 0.

4 = 0 + C

So C = 4.

C = 4

Put C back.

The answer: y = x² + x + 4.

y = x² + x + 4

4 Check

Let's check: the derivative of our answer is 2x + 1.

The right side again.

Right!

y′ = 2x + 1

And at x = 0, y is 4: the starting value.

Right!

y(0) = 4

Let's look back.

First: the idea.

Then: integrate both sides.

Then: use the starting value.

Last: check.

Answer