Second-Order Initial Value Problems
Chapter Seventy-Seven
Syllabus topic Module 1, "1.3 Inverse Laplace Transform"
Pages 170 to 171 of 303
In one line
Two initial values go in, a quadratic in s comes out, and its discriminant decides whether the answer decays, oscillates, or sits on the boundary.
The transform row
L{diff(y(t), (t, 2))} = s^2 Y(s) - s y(0) - diff(y(0), t)
The s sits on y(0), not on y prime of 0. Getting that the wrong way round is the commonest error in this whole chapter.
Worked: real distinct roots
Solve y double prime minus 5y prime plus 6y equals 0, with y(0) = 1 and y prime of 0 = 0.
Transforming: (s squared Y minus s minus 0) minus 5(sY minus 1) plus 6Y equals 0.
Collecting the Y terms: Y(s squared minus 5s plus 6) equals s minus 5.
So Y equals (s minus 5) over (s squared minus 5s plus 6), and the denominator factorises as (s minus 2)(s minus 3).
By the cover-up rule: at s = 2 the numerator is minus 3 and the other factor is minus 1, so the constant is 3. At s = 3 the numerator is minus 2 and the other factor is 1, so the constant is minus 2.
(s - 5)/((s - 2)(s - 3)) = 3/(s - 2) - 2/(s - 3)
L⁻¹{(s - 5)/((s - 2)(s - 3))} = 3 e^(2t) - 2 e^(3t)
d2y/dx2 - 5 dy/dx + 6y = 0
y = 3 e^(2x) - 2 e^(3x)
Both conditions check: at x = 0 the value is 3 minus 2, which is 1, and the derivative is 6 minus 6, which is 0.
Worked: complex roots
Solve y double prime plus 2y prime plus 5y equals 0, with y(0) = 0 and y prime of 0 = 4.
Transforming: (s squared Y minus 0 minus 4) plus 2(sY minus 0) plus 5Y equals 0.
Collecting: Y(s squared plus 2s plus 5) equals 4, so Y equals 4 over (s squared plus 2s plus 5).
The discriminant is 4 minus 20, which is negative, so complete the square: the denominator is (s plus 1) squared plus 4.
L⁻¹{4/((s + 1)^2 + 4)} = 2 e^(-t) sin(2t)
d2y/dx2 + 2 dy/dx + 5y = 0
y = 2 e^(-x) sin(2x)
At x = 0 the value is 0, and the derivative at 0 is 4, both as required. The answer is a decaying oscillation, which is what a negative discriminant always gives.
Worked: a repeated root
Solve y double prime plus 4y prime plus 4y equals 0, with y(0) = 1 and y prime of 0 = 0.
Transforming and collecting: Y(s squared plus 4s plus 4) equals s plus 4, and the denominator is (s plus 2) squared.
Write the numerator in terms of s plus 2: it is (s plus 2) plus 2.
Second-Order Initial Value Problems
L⁻¹{(s + 4)/(s + 2)^2} = e^(-2t) + 2 t e^(-2t)
d2y/dx2 + 4 dy/dx + 4y = 0
y = e^(-2x) + 2 x e^(-2x)
The t in the answer is the repeated root, exactly as the partial-fractions chapter said.
Worked: with a forcing term
Solve y double prime plus y equals 1, with y(0) = 0 and y prime of 0 = 0.
Transforming: s squared Y plus Y equals one over s, so Y equals one over s(s squared plus 1).
L⁻¹{1/(s(s^2 + 1))} = 1 - cos(t)
d2y/dx2 + y = 1
y = 1 - cos(x)
Worked: resonance
Solve y double prime plus 4y equals sin 2t, with y(0) = 0 and y prime of 0 = 0.
The forcing has exactly the frequency the system oscillates at. Transforming: s squared Y plus 4Y equals 2 over (s squared plus 4), so Y equals 2 over (s squared plus 4) squared.
That is the repeated quadratic, which the convolution chapter inverted.
L⁻¹{2/(s^2 + 4)^2} = (sin(2t) - 2 t cos(2t))/8
d2y/dx2 + 4y = sin(2x)
y = (sin(2x) - 2 x cos(2x))/8
The t cos 2t term grows without limit. That is resonance: driving a system at its own frequency makes the amplitude build up for ever, which is why soldiers break step on a bridge.
The discriminant decides everything
| Discriminant of the quadratic in s | Roots | Answer |
|---|---|---|
| positive | real and distinct | two exponentials |
| zero | repeated | an exponential and t times it |
| negative | complex pair | a decaying or growing oscillation |
Work out the discriminant before anything else and you know what the answer will look like. Module 2's chapters on the auxiliary equation say the same thing in the other language, and it is worth noticing that they are the same three cases.
Check yourself
d2y/dx2 - y = 0
y = C1 e^x + C2 e^(-x)
d2y/dx2 + 9y = 0
y = C1 cos(3x) + C2 sin(3x)
d2y/dx2 + 2 dy/dx + y = 0
y = C1 e^(-x) + C2 x e^(-x)
Those are the general solutions of the three cases. A transform problem would fix C1 and C2 from the initial values; here they are left free so that the checker proves the solution for every value of both constants, which is a stronger statement.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.