Worked examples
The question
Question: solve dy/dx = 6x².
dydx = 6x2
1 The idea
dy/dx is only about x.
So y is a function whose derivative is 6x²: an antiderivative.
y′ = 6x2: y is a function whose derivative is 6x2
2 Integrate both sides
Integrate both sides.
The left side, the integral of dy/dx, is just y.
y = ∫ 6x2 dx
The power rule backwards: add 1 to the power, divide by the new power.
∫ xn dx = xn + 1n + 1
Each term by its rule, and one + C: the number we cannot see after a derivative.
y = 6 × x33 + C
Tidy the numbers: y = 2x³ + C.
y = 2x3 + C
Careful!
Don't forget + C.
Every number C gives another solution.
Careful: not y = 2x3 ✗
The answer: y = 2x³ + C.
C can be any number: one solution for each C.
y = 2x3 + C
3 Check
Let's check: the derivative of our answer (the C goes, its derivative is 0) is 6x².
The right side again.
Right!
y′ = 6x2
Let's look back.
First: the idea.
Then: integrate both sides.
Last: check.
▶ Watch it solved
The question
Question: solve y′ = 4x³ − 2x, with y(1) = 5.
y′ = 4x3 − 2x
y(1) = 5 is the starting value: when x is 1, y is 5.
y(1) = 5
1 The idea
y′ is only about x.
So y is a function whose derivative is 4x³ − 2x: an antiderivative.
y′ = 4x3 − 2x: y is a function whose derivative is 4x3 − 2x
2 Integrate both sides
Integrate both sides.
The left side, the integral of y′, is just y.
y = ∫ (4x3 − 2x) dx
The power rule backwards: add 1 to the power, divide by the new power.
∫ xn dx = xn + 1n + 1
Each term by its rule, and one + C: the number we cannot see after a derivative.
y = 4 × x44 − 2 × x22 + C
Tidy the numbers: y = x⁴ − x² + C.
y = x4 − x2 + C
Careful!
Don't forget + C.
Every number C gives another solution, and we need C to use the starting value.
Careful: not y = x4 − x2 ✗
3 Use the starting value
Put in x = 1 and y = 5.
5 = (1)4 − (1)2 + C
Work out the numbers: 0.
5 = 0 + C
So C = 5.
C = 5
Put C back.
The answer: y = x⁴ − x² + 5.
y = x4 − x2 + 5
4 Check
Let's check: the derivative of our answer is 4x³ − 2x.
The right side again.
Right!
y′ = 4x3 − 2x
And at x = 1, y is 5: the starting value.
Right!
y(1) = 5
Let's look back.
First: the idea.
Then: integrate both sides.
Then: use the starting value.
Last: check.
▶ Watch it solved
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 = xn + 1n + 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 y = 3sin(2x) − 2x2 + 2
▶ Watch it solved
