Worked examples
The question
Question: check if y = e^(3x) solves y′ = 3y.
y = e^(3x)
y′ = 3y is a differential equation: an equation with a derivative in it.
y′ = 3y
Its answer is not a number.
It is a function y, and it must make the equation true for every x.
differential equation: an equation with a derivative
its answer: a function y, true for every x
1 Find the derivatives
The derivative of e^(3x) is 3e^(3x).
y′ = 3e^(3x)
The rules we use: the chain rule brings the number in front down.
(e^(kx))′ = ke^(kx)
Careful!
The chain rule!
Times the inside's slope, 3.
Careful: not (e^(3x))′ = e^(3x) ✗
2 Put them in
Now put y and y′ into the equation, in place of the letters.
3e^(3x) = 3 × e^(3x)
Multiply out and add like terms.
3e^(3x) = 3e^(3x)
3 Compare the sides
Both sides are the same, for every x.
So y = e^(3x) is a solution.
= yes: y = e^(3x) is a solution
4 Check
Let's check with one number, x = 1.
A calculator gives 60.26 on both sides.
Right!
x = 1: left ≈ 60.26, right ≈ 60.26
Let's look back.
First: find the derivatives.
Then: put them in.
Then: compare the sides.
Last: check.
Answer = yes: y = e^(3x) is a solution
▶ Watch it solved
The question
Question: check if y = sin(2x) solves y″ + 4y = 0.
y = sin(2x)
y″ + 4y = 0 is a differential equation: an equation with a derivative in it.
y″ + 4y = 0
Its answer is not a number.
It is a function y, and it must make the equation true for every x.
differential equation: an equation with a derivative
its answer: a function y, true for every x
1 Find the derivatives
The derivative of sin(2x) is 2cos(2x).
y′ = 2cos(2x)
The rules we use: the chain rule brings the number in front down.
(sin(bx))′ = b cos(bx)
Careful!
The chain rule!
Times the inside's slope, 2.
Careful: not (sin(2x))′ = cos(2x) ✗
Once more: y″ is the derivative of y′.
It is −4sin(2x).
y″ = −4sin(2x)
2 Put them in
Now put y, y′ and y″ into the equation, in place of the letters.
−4sin(2x) + 4 × sin(2x) = 0
Multiply out and add like terms.
On the left, everything cancels: 0.
0 = 0
3 Compare the sides
Both sides are the same, for every x.
So y = sin(2x) is a solution.
= yes: y = sin(2x) is a solution
4 Check
Let's check with one number, x = 1.
A calculator gives 0 on both sides.
Right!
x = 1: left ≈ −3.64 + 3.64 = 0, right ≈ 0
Let's look back.
First: find the derivatives.
Then: put them in.
Then: compare the sides.
Last: check.
Answer = yes: y = sin(2x) is a solution
▶ Watch it solved
The question
Question: check if y = xe^(2x) solves y″ − 4y′ + 4y = 0.
y = xe^(2x)
y″ − 4y′ + 4y = 0 is a differential equation: an equation with a derivative in it.
y″ − 4y′ + 4y = 0
Its answer is not a number.
It is a function y, and it must make the equation true for every x.
differential equation: an equation with a derivative
its answer: a function y, true for every x
1 Find the derivatives
The derivative of xe^(2x) is e^(2x) + 2xe^(2x).
y′ = e^(2x) + 2xe^(2x)
The rules we use.
A product needs the product rule.
(e^(kx))′ = ke^(kx)
(xⁿ)′ = nxⁿ⁻¹
(uv)′ = u′v + uv′
Careful!
The chain rule!
Times the inside's slope, 2.
Careful: not (e^(2x))′ = e^(2x) ✗
Once more: y″ is the derivative of y′.
It is 4e^(2x) + 4xe^(2x).
y″ = 4e^(2x) + 4xe^(2x)
2 Put them in
Now put y, y′ and y″ into the equation, in place of the letters.
4e^(2x) + 4xe^(2x) − 4(e^(2x) + 2xe^(2x)) + 4 × xe^(2x) = 0
Multiply out and add like terms.
On the left, everything cancels: 0.
0 = 0
3 Compare the sides
Both sides are the same, for every x.
So y = xe^(2x) is a solution.
= yes: y = xe^(2x) is a solution
4 Check
Let's check with one number, x = 1.
A calculator gives 0 on both sides.
Right!
x = 1: left ≈ 59.11 − 88.67 + 29.56 = 0, right ≈ 0
Let's look back.
First: find the derivatives.
Then: put them in.
Then: compare the sides.
Last: check.
Answer = yes: y = xe^(2x) is a solution
▶ Watch it solved
