Worked examples
The question
Question: solve y″ − 5y′ + 6y = 0.
y″ − 5y′ + 6y = 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) − 5re^(rx) + 6e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 − 5r + 6) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² − 5r + 6 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 − 5r + 6 = 0
3 Solve for r
Two numbers that multiply to 6 and add to −5: −2 and −3.
−2 × (−3) = 6, −2 + (−3) = −5
So it factors: (r − 2)(r − 3) = 0.
(r − 2)(r − 3) = 0
Each bracket can be 0: r = 2 or r = 3.
r = 2 or r = 3
Careful!
(r − 2) is 0 when r is 2: the opposite sign of the number in the bracket.
Careful: not r = −2, −3 ✗
4 Write the solution
Each root gives a solution e^(rx).
Any mix of the two works, so we add them with two numbers, C₁ and C₂.
two roots r₁, r₂: C₁e^(r₁x) + C₂e^(r₂x)
So y = C₁e^(2x) + C₂e^(3x).
y = C1e^(2x) + C2e^(3x)
5 Check
Let's check: each root must make r² − 5r + 6 = 0.
r = 2 gives 0.
r = 2: 2² − 5 × 2 + 6 = 0
And r = 3 gives 0.
Right!
r = 3: 3² − 5 × 3 + 6 = 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 = C1e^(2x) + C2e^(3x)
▶ Watch it solved
The question
Question: solve y″ + 2y′ − 8y = 0.
y″ + 2y′ − 8y = 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) + 2re^(rx) − 8e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 + 2r − 8) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² + 2r − 8 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 + 2r − 8 = 0
3 Solve for r
Two numbers that multiply to −8 and add to 2: 4 and −2.
4 × (−2) = −8, 4 + (−2) = 2
So it factors: (r + 4)(r − 2) = 0.
(r + 4)(r − 2) = 0
Each bracket can be 0: r = −4 or r = 2.
r = −4 or r = 2
Careful!
(r + 4) is 0 when r is −4: the opposite sign of the number in the bracket.
Careful: not r = 4, −2 ✗
4 Write the solution
Each root gives a solution e^(rx).
Any mix of the two works, so we add them with two numbers, C₁ and C₂.
two roots r₁, r₂: C₁e^(r₁x) + C₂e^(r₂x)
So y = C₁e^(−4x) + C₂e^(2x).
y = C1e^(−4x) + C2e^(2x)
5 Check
Let's check: each root must make r² + 2r − 8 = 0.
r = −4 gives 0.
r = −4: (−4)² + 2 × (−4) − 8 = 0
And r = 2 gives 0.
Right!
r = 2: 2² + 2 × 2 − 8 = 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 = C1e^(−4x) + C2e^(2x)
▶ Watch it solved
The question
Question: solve y″ − 5y′ + 6y = 0, with y(0) = 2 and y′(0) = 5.
y″ − 5y′ + 6y = 0
The starting values: at x = 0, y is 2 and its slope y′ is 5.
y(0) = 2, y′(0) = 5
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) − 5re^(rx) + 6e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 − 5r + 6) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² − 5r + 6 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 − 5r + 6 = 0
3 Solve for r
Two numbers that multiply to 6 and add to −5: −2 and −3.
−2 × (−3) = 6, −2 + (−3) = −5
So it factors: (r − 2)(r − 3) = 0.
(r − 2)(r − 3) = 0
Each bracket can be 0: r = 2 or r = 3.
r = 2 or r = 3
Careful!
(r − 2) is 0 when r is 2: the opposite sign of the number in the bracket.
Careful: not r = −2, −3 ✗
4 Write the solution
Each root gives a solution e^(rx).
Any mix of the two works, so we add them with two numbers, C₁ and C₂.
two roots r₁, r₂: C₁e^(r₁x) + C₂e^(r₂x)
So y = C₁e^(2x) + C₂e^(3x).
y = C1e^(2x) + C2e^(3x)
5 Use the starting values
We need y′ too: the derivative of our y.
y′ = 2C1e^(2x) + 3C2e^(3x)
at x = 0: e^(0) = 1,
Put x = 0 into y: C₁ + C₂ = 2.
① C1 + C2 = 2
Put x = 0 into y′: 2C₁ + 3C₂ = 5.
② 2C1 + 3C2 = 5
From ①: C₁ = 2 − C₂.
C1 = 2 − C2
Put it into ②.
2(2 − C2) + 3C2 = 5
Open the bracket and collect: 2 × 2 moves over.
C2 = 1
So C₂ = 1, and C₁ = 1.
C2 = 1
Put them in: y = e^(2x) + e^(3x).
y = e^(2x) + e^(3x)
6 Check
Let's check: each root must make r² − 5r + 6 = 0.
r = 2 gives 0.
r = 2: 2² − 5 × 2 + 6 = 0
And r = 3 gives 0.
Right!
r = 3: 3² − 5 × 3 + 6 = 0
And the starting values: y(0) = 2 and y′(0) = 5.
Right!
y(0) = 2, y′(0) = 5
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^(2x) + e^(3x)
▶ Watch it solved
