When the Exponential Rule Fails, and What To Do
Chapter One Hundred Twenty-Seven
Syllabus topic Module 2, "2.3 Linear Differential Equations with Constant Coefficients"
Pages 282 to 283 of 303
In one line
If f(a) is zero, divide out the factor and bring in a factor of x, once for each time a is a root.
Why it fails
The rule says the PI is e to the ax over f(a). If f(a) is zero, that is a division by zero and the rule says nothing.
And f(a) being zero means a is a root of the auxiliary equation, so e to the ax is part of the complementary function. The driving term is one of the system's own natural motions, which is resonance.
So the failure is not an algebraic accident. It is the mathematics telling you that something physically important is happening.
The repair
If (D minus a) is a factor of f(D) exactly once, write f(D) as (D minus a)g(D) with g(a) not zero.
(1/f(D)) e^(a x) = x e^(a x)/g(a)
If (D minus a) is a factor k times, the answer carries x to the power k, divided by k factorial and by the rest of the operator evaluated at a.
(1/f(D)) e^(a x) = x^k e^(a x)/(factorial(k) g(a))
Where the x comes from
Apply the shift relation of the D-laws chapter. One over (D minus a) acting on e to the ax equals e to the ax times one over D acting on 1, and one over D is integration, so that is e to the ax times x.
integrate(1, x) = x
Applied twice, one over (D minus a) squared gives e to the ax times the integral of x, which is x squared over 2. That is where the k factorial comes from.
integrate(x, x) = x^2/2
integrate(integrate(x, x), x) = x^3/6
Worked, a simple root
Solve y'' minus 5y' plus 6y = e to the 2x.
f(D) is (D minus 2)(D minus 3), so f(2) is zero: the rule has failed.
Write g(D) as (D minus 3), so g(2) is minus 1.
The PI is therefore x e to the 2x over minus 1, that is minus x e to the 2x.
d2y/dx2 - 5 dy/dx + 6y = e^(2x)
y = C1 e^(2x) + C2 e^(3x) - x e^(2x)
Notice that the PI contains e to the 2x, which is also in the CF. That is fine: what makes the PI a PI is the x in front of it, and no choice of C1 can produce x e to the 2x.
Worked, a repeated root
Solve y'' minus 4y' plus 4y = e to the 2x.
f(D) is (D minus 2) squared, so 2 is a root twice: k = 2, and g(D) is 1, so g(2) is 1.
The PI is x squared e to the 2x over 2 factorial, which is x squared e to the 2x over 2.
When the Exponential Rule Fails, and What To Do
The CF for (D minus 2) squared is C1 e to the 2x plus C2 x e to the 2x, by the repeated-roots chapter, so the complete solution is that plus the PI.
d2y/dx2 - 4 dy/dx + 4y = e^(2x)
y = C1 e^(2x) + C2 x e^(2x) + x^2 e^(2x)/2
Three terms, and the powers of x run 0, 1, 2: two from the CF and the third from the PI. The PI always carries the next power of x after the CF's highest, which is a useful check.
Worked, a constant with a factor of D
A constant right-hand side fails the rule when f(0) is zero, which happens when f(D) has a factor of D.
Solve y'' plus 3y' = 6.
f(D) is D(D plus 3), so f(0) is zero. Here a = 0, k = 1, and g(D) is (D plus 3), so g(0) is 3.
The PI is 6 x over 3, which is 2x.
d2y/dx2 + 3 dy/dx = 6
y = C1 + C2 e^(-3x) + 2 x
So a constant right-hand side gives a PI linear in x when the operator has a factor of D. That is a standard question and it catches students who reach for a constant PI automatically.
The physical meaning, stated plainly
Drive a system at one of its own natural frequencies and the response grows. The x in front of the exponential is that growth.
For a decaying exponential the x eventually loses to the decay, so the response rises and then falls. For a growing one, or for a pure oscillation, the x makes the response grow without limit. That is why soldiers break step on a bridge, and why a wine glass shatters at the right note.
Check yourself
d2y/dx2 - y = e^x
y = C1 e^x + C2 e^(-x) + x e^x/2
d2y/dx2 - 2 dy/dx + y = e^x
y = C1 e^x + C2 x e^x + x^2 e^x/2
d2y/dx2 + dy/dx = 2
y = C1 + C2 e^(-x) + 2 x
For the first, f(D) is (D minus 1)(D plus 1), so g(1) is 2 and the PI is x e to the x over 2. For the second, 1 is a double root, so the PI carries x squared over 2 factorial.
The rest of this subject
These notes are cut from the University's printed syllabus. Open the syllabus itself for the same subject.