Worked examples
The question
Question: solve y″ + 9y = 0.
y″ + 9y = 0
second order, linear, constant numbers, 0 on the right
1 Try y = e to the rx
The derivatives of e to the rx are e to the rx again, times r.
So we try y = e^(rx), and look for the right r.
y = e^(rx): each derivative keeps the same shape
y = e^(rx)
Its derivative: r e^(rx).
y′ = re^(rx)
Once more: r² e^(rx).
y″ = r2e^(rx)
2 The characteristic equation
Put them into the equation.
r2e^(rx) + 9e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 + 9) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² + 9 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 + 9 = 0
3 Solve for r
It does not factor nicely, so we use the formula, with a = 1, b = 0, c = 9.
r = −b ± √b² − 4ac2a
Put the numbers in.
r = 0 ± √0² − 4 × 1 × 92 × 1
Under the root: −36.
r = 0 ± √−362
A minus under the root: √(−36) = 6i, because i = √(−1).
√−36 = √36 × √−1 = 6i
r = 0 ± 6i2
Divide both parts by 2: r = ±3i.
Two complex roots.
r = ±3i
Careful!
Use the root, 3, not the number under it.
Careful: not cos(9x) ✗
4 Write the solution
Complex roots give cosines and sines.
The real part 0 goes in the e, the 3 goes inside cos and sin.
α ± βi: e^(αx)(C₁cos βx + C₂sin βx)
So y = C₁cos(3x) + C₂sin(3x).
y = C1cos(3x) + C2sin(3x)
5 Check
Let's check one root, 3i.
Square it, with i² = −1: −9.
r = 3i: r² = −9
Into r² + 9: the numbers cancel and the i parts cancel.
0.
Right!
−9 + 9 = 0
Let's look back.
First: try y = e to the rx.
Then: the characteristic equation.
Then: solve for r.
Then: write the solution.
Last: check.
Answer y = C1cos(3x) + C2sin(3x)
▶ Watch it solved
The question
Question: solve y″ − 4y′ + 5y = 0.
y″ − 4y′ + 5y = 0
second order, linear, constant numbers, 0 on the right
1 Try y = e to the rx
The derivatives of e to the rx are e to the rx again, times r.
So we try y = e^(rx), and look for the right r.
y = e^(rx): each derivative keeps the same shape
y = e^(rx)
Its derivative: r e^(rx).
y′ = re^(rx)
Once more: r² e^(rx).
y″ = r2e^(rx)
2 The characteristic equation
Put them into the equation.
r2e^(rx) − 4re^(rx) + 5e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 − 4r + 5) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² − 4r + 5 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 − 4r + 5 = 0
3 Solve for r
It does not factor nicely, so we use the formula, with a = 1, b = −4, c = 5.
r = −b ± √b² − 4ac2a
Put the numbers in.
r = 4 ± √(−4)² − 4 × 1 × 52 × 1
Under the root: −4.
r = 4 ± √−42
A minus under the root: √(−4) = 2i, because i = √(−1).
√−4 = √4 × √−1 = 2i
r = 4 ± 2i2
Divide both parts by 2: r = 2 ± i.
Two complex roots.
r = 2 ± i
Careful!
Don't lose the real part, 2.
It gives e^(2x) in front.
Careful: not y = C₁cos(x) + C₂sin(x) ✗
4 Write the solution
Complex roots give cosines and sines.
The real part 2 goes in the e, the 1 goes inside cos and sin.
α ± βi: e^(αx)(C₁cos βx + C₂sin βx)
So y = e^(2x)(C₁cos x + C₂sin x).
y = e^(2x)(C1cos x + C2sin x)
5 Check
Let's check one root, 2 + i.
Square it, with i² = −1: 3 + 4i.
r = 2 + i: r² = 3 + 4i
Into r² − 4r + 5: the numbers cancel and the i parts cancel.
0.
Right!
3 + 4i − 4(2 + i) + 5 = 0
Let's look back.
First: try y = e to the rx.
Then: the characteristic equation.
Then: solve for r.
Then: write the solution.
Last: check.
Answer y = e^(2x)(C1cos x + C2sin x)
▶ Watch it solved
The question
Question: solve y″ + 2y′ + 5y = 0, with y(0) = 1 and y′(0) = 3.
y″ + 2y′ + 5y = 0
The starting values: at x = 0, y is 1 and its slope y′ is 3.
y(0) = 1, y′(0) = 3
second order, linear, constant numbers, 0 on the right
1 Try y = e to the rx
The derivatives of e to the rx are e to the rx again, times r.
So we try y = e^(rx), and look for the right r.
y = e^(rx): each derivative keeps the same shape
y = e^(rx)
Its derivative: r e^(rx).
y′ = re^(rx)
Once more: r² e^(rx).
y″ = r2e^(rx)
2 The characteristic equation
Put them into the equation.
r2e^(rx) + 2re^(rx) + 5e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 + 2r + 5) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² + 2r + 5 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 + 2r + 5 = 0
3 Solve for r
It does not factor nicely, so we use the formula, with a = 1, b = 2, c = 5.
r = −b ± √b² − 4ac2a
Put the numbers in.
r = −2 ± √2² − 4 × 1 × 52 × 1
Under the root: −16.
r = −2 ± √−162
A minus under the root: √(−16) = 4i, because i = √(−1).
√−16 = √16 × √−1 = 4i
r = −2 ± 4i2
Divide both parts by 2: r = −1 ± 2i.
Two complex roots.
r = −1 ± 2i
Careful!
Don't lose the real part, −1.
It gives e^(−x) in front.
Careful: not y = C₁cos(2x) + C₂sin(2x) ✗
4 Write the solution
Complex roots give cosines and sines.
The real part −1 goes in the e, the 2 goes inside cos and sin.
α ± βi: e^(αx)(C₁cos βx + C₂sin βx)
So y = e^(−x)(C₁cos(2x) + C₂sin(2x)).
y = e^(−x)(C1cos(2x) + C2sin(2x))
5 Use the starting values
We need y′ too: the derivative of our y.
y′ = −C1e^(−x)cos(2x) − 2C1e^(−x)sin(2x) − C2e^(−x)sin(2x) + 2C2e^(−x)cos(2x)
at x = 0: e^(0) = 1, cos 0 = 1, sin 0 = 0
Put x = 0 into y: C₁ = 1.
So C₁ is known at once.
① C1 = 1
Put x = 0 into y′: −C₁ + 2C₂ = 3.
② −C1 + 2C2 = 3
Put C₁ into ②.
−1 × 1 + 2C2 = 3
So C₂ = 2.
C2 = 2
Put them in: y = e^(−x)(cos(2x) + 2sin(2x)).
y = e^(−x)(cos(2x) + 2sin(2x))
6 Check
Let's check one root, −1 + 2i.
Square it, with i² = −1: −3 − 4i.
r = −1 + 2i: r² = −3 − 4i
Into r² + 2r + 5: the numbers cancel and the i parts cancel.
0.
Right!
−3 − 4i + 2(−1 + 2i) + 5 = 0
And the starting values: y(0) = 1 and y′(0) = 3.
Right!
y(0) = 1, y′(0) = 3
Let's look back.
First: try y = e to the rx.
Then: the characteristic equation.
Then: solve for r.
Then: write the solution.
Then: use the starting values.
Last: check.
Answer y = e^(−x)(cos(2x) + 2sin(2x))
▶ Watch it solved
