Worked examples
The question
Question: solve y″ + 4y′ + 4y = 0.
y″ + 4y′ + 4y = 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) + 4e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 + 4r + 4) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² + 4r + 4 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 + 4r + 4 = 0
3 Solve for r
Two numbers that multiply to 4 and add to 4: 2 and 2, the same number twice.
2 × 2 = 4, 2 + 2 = 4
So it is a square: (r + 2)² = 0.
(r + 2)2 = 0
Only one root: r = −2, twice.
r = −2 (twice)
4 Write the solution
One root gives only one solution.
The second one is the same times x.
(Put it in: it works.)
one root twice: C₁e^(rx) + C₂xe^(rx)
Careful!
That is only one solution twice: (C₁ + C₂)e^(−2x).
The second one needs an x.
Careful: not y = C1e^(−2x) + C2e^(−2x) ✗
So y = C₁e^(−2x) + C₂xe^(−2x).
y = C1e^(−2x) + C2xe^(−2x)
5 Check
Let's check the new part, xe^(−2x): its derivatives, by the product rule.
y = xe^(−2x): y′ = e^(−2x) − 2xe^(−2x), y″ = −4e^(−2x) + 4xe^(−2x)
Put them in: everything cancels, 0.
Right!
put in: −4e^(−2x) + 4xe^(−2x) + 4(e^(−2x) − 2xe^(−2x)) + 4 × xe^(−2x) = 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) + C2xe^(−2x)
▶ Watch it solved
The question
Question: solve 4y″ + 4y′ + y = 0.
4y″ + 4y′ + y = 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.
4r2e^(rx) + 4re^(rx) + e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(4r2 + 4r + 1) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
4r² + 4r + 1 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
4r2 + 4r + 1 = 0
3 Solve for r
It does not factor nicely, so we use the formula, with a = 4, b = 4, c = 1.
r = −b ± √b² − 4ac2a
Put the numbers in.
r = −4 ± √4² − 4 × 4 × 12 × 4
Under the root: 0.
r = −4 ± √08
Under the root is 0, so plus and minus give the same: r = −1/2, twice.
r = −1/2 (twice)
4 Write the solution
One root gives only one solution.
The second one is the same times x.
(Put it in: it works.)
one root twice: C₁e^(rx) + C₂xe^(rx)
Careful!
That is only one solution twice: (C₁ + C₂)e^(−x/2).
The second one needs an x.
Careful: not y = C1e^(−x/2) + C2e^(−x/2) ✗
So y = C₁e^(−x/2) + C₂xe^(−x/2).
y = C1e^(−x/2) + C2xe^(−x/2)
5 Check
Let's check the new part, xe^(−x/2): its derivatives, by the product rule.
y = xe^(−x/2): y′ = e^(−x/2) − 12xe^(−x/2), y″ = −e^(−x/2) + 14xe^(−x/2)
Put them in: everything cancels, 0.
Right!
put in: 4(−e^(−x/2) + 14xe^(−x/2)) + 4(e^(−x/2) − 12xe^(−x/2)) + xe^(−x/2) = 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^(−x/2) + C2xe^(−x/2)
▶ Watch it solved
The question
Question: solve y″ − 6y′ + 9y = 0, with y(0) = 2 and y′(0) = 7.
y″ − 6y′ + 9y = 0
The starting values: at x = 0, y is 2 and its slope y′ is 7.
y(0) = 2, y′(0) = 7
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) − 6re^(rx) + 9e^(rx) = 0
Every term has e^(rx): take it out.
e^(rx)(r2 − 6r + 9) = 0
e to a power is never 0.
So the bracket must be 0.
e^(rx) is never 0
r² − 6r + 9 = 0: the characteristic equation.
y″ became r², y′ became r, y became 1.
r2 − 6r + 9 = 0
3 Solve for r
Two numbers that multiply to 9 and add to −6: −3 and −3, the same number twice.
−3 × (−3) = 9, −3 + (−3) = −6
So it is a square: (r − 3)² = 0.
(r − 3)2 = 0
Only one root: r = 3, twice.
r = 3 (twice)
4 Write the solution
One root gives only one solution.
The second one is the same times x.
(Put it in: it works.)
one root twice: C₁e^(rx) + C₂xe^(rx)
Careful!
That is only one solution twice: (C₁ + C₂)e^(3x).
The second one needs an x.
Careful: not y = C1e^(3x) + C2e^(3x) ✗
So y = C₁e^(3x) + C₂xe^(3x).
y = C1e^(3x) + C2xe^(3x)
5 Use the starting values
We need y′ too: the derivative of our y.
y′ = 3C1e^(3x) + C2e^(3x) + 3C2xe^(3x)
at x = 0: e^(0) = 1, x = 0
Put x = 0 into y: C₁ = 2.
So C₁ is known at once.
① C1 = 2
Put x = 0 into y′: 3C₁ + C₂ = 7.
② 3C1 + C2 = 7
Put C₁ into ②.
3 × 2 + C2 = 7
So C₂ = 1.
C2 = 1
Put them in: y = 2e^(3x) + xe^(3x).
y = 2e^(3x) + xe^(3x)
6 Check
Let's check the new part, xe^(3x): its derivatives, by the product rule.
y = xe^(3x): y′ = e^(3x) + 3xe^(3x), y″ = 6e^(3x) + 9xe^(3x)
Put them in: everything cancels, 0.
Right!
put in: 6e^(3x) + 9xe^(3x) − 6(e^(3x) + 3xe^(3x)) + 9 × xe^(3x) = 0
And the starting values: y(0) = 2 and y′(0) = 7.
Right!
y(0) = 2, y′(0) = 7
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 = 2e^(3x) + xe^(3x)
▶ Watch it solved
