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Math 3D - Additional “Judicious Guessing” Examples Paul Macklin February 4, 2005
The method of judicious guessing can be tricky to get the hang of, but it is rewarding. It can save you a lot of work when compared to variation of parameters, even in the longer examples.
Here, I hope to help clarify some additional cases of “judicious guessing”. I’m going to work several of the odd exercises in your text (page 164), as well as problem 10, the “stumper problem”. Many thanks to those who contributed ideas on how to solve that problem!
Exercises Find a particular solution to each of the following equations.
- y′′^ + 3y = t^3 − 1.
Solution: Because the right-hand side is a polynomial, we expect ψ to be of the form of a polynomial. Next, notice that the solutions to the homogeneous problem are cos(
3 t) and sin(
3 t), and so neither t^3 nor 1 is a solution to the homogeneous DE. So, we only expect to need a 3rd^ degree polynomial:
ψ(t) = a + bt + ct^2 + dt^3 ψ′(t) = b + 3ct + 3dt^2 ψ′′(t) = 3 c + 6dt,
and so
ψ′′^3 c + 6dt +3ψ 3 a + 3bt + 3ct^2 + 3dt^3 t^3 (1) + t^2 (0) + t(0) + 1(−1) = t^3 (3d) + t^2 (3c) + t(3b + 6d) + 1(3c + 2a).
By equating coefficients of these polynomials, we get
d =^1 3 , c = 0, b = − 2 3 , a = − 1 3
and so ψ(t) = −
t +
t^3.
- y′′^ − y = t^2 et
Solution: Because the right-hand side is a polynomial times an exponential, we expect a particular solution of the form
ψ(t) = v(t)et ψ′(t) = v′et^ + vet ψ′′(t) = v′′et^ + 2v′et^ + vet
where v is a polynomial of degree at least 2. Let’s plug this in to get a better feeling for what degree v needs to be.
ψ′′^ et(v′′^ + 2v′^ + v) −ψ et(v) et
t^2 (1) + t(0) + 1(0)
= et
v′′^ + 2v′
Notice that if the degree of v is m, then the degree of v′′^ + 2v′^ is m − 1. Because we need this to be a second-degree polynomial, we need m − 1 = 2, i.e., m = 3. Therefore, ψ = v(t)et^ = et(a + bt + ct^2 + dt^3 ). If we plug in this refined guess (into (1), we get
et(t^2 ) = et
2 c + 6dt + 2b + 4ct + 6dt^2
= et
t^2 (6d) + t(6d + 4c) + 1(2c + 2b)
By equating coefficients, we have d =^1 6 , c = − 1 4 , b =^1 4
and a is undetermined. Therefore, we get a whole family of solutions
ψ = et
a +
t −
t^2 +
t^3
We might as well pick the simplest solution and set a = 0.
ψ = et
t −
t^2 +
t^3
. = tet
t +
t^2
- y′′^ + 2y′^ + y = e−t
Solution: Because the right-hand side is an exponential, we might expect a particular solution of the form ce−t. However, we see that both e−t^ and te−t^ are solutions to the homogeneous equation. So, instead we expect a particular solution of the form ψ(t) = v(t)e−t ψ′(t) = v′e−t^ − ve−t ψ′′(t) = v′′e−t^ − 2 v′e−t^ + ve−t where v is a polynomial. Let’s plug this in to get a better feeling for what degree v needs to be. ψ′′^ e−t(v′′^ − 2 v′^ + v) +2ψ′^ e−t(2v′^ − 2 v) −ψ e−t(v) e−t
= e−t
v′′
Notice that if the degree of v is m, then the degree of v′′^ is m − 2. Because we need this to be a constant polynomial, we need m − 2 = 0, i.e., m = 2. Therefore, ψ = v(t)e−t^ = e−t(a + bt + ct^2 ). Notice that since e−t^ and te−t^ are solutions to the homogeneous equation, we have L(ψ) = aL(e−t) + bL(te−t) + cL(t^2 e−t) = 0 + 0 + cL(t^2 e−t). Thus, our choices for a and b are irrelevant, as they cannot contribute towards a nonzero righthand side. Thus, we can set a = b = 0 right away. If this is the case, and if we plug in this refined guess, we get e−t(1) = et
2 c
By equating coefficients, we have c =
Now, if L(y) = y′′^ + 4y is the linear operator for the LHS, then L(Ψ) = te^2 it^ = t cos(2t) + it sin(2t). Therefore, by equating imaginary and real parts (and by using rules of linear operators),
L
t cos 2t + 2t^2 sin(2t)
= t cos(2t),
and L
t sin(2t) − 2 t^2 cos(2t)
= t sin(2t),
Because our original RHS was t sin(2t), we choose
ψ(t) =
t sin(2t) − 2 t^2 cos(2t)
- y′′^ − 2 y′^ + 5y = 2 cos^2 t
Solution: First, recall the half-angle identity:
cos^2 t = 1 + cos(2t) 2
Therefore, the DE is equivalent to L(y) = y′′^ − 2 y′^ + 5y = 1 + cos(2t). Let us consider the related DE L(y) = 1 + e^2 it. Notice first that if y = 15 , then y′′^ − 2 y′^ + 5y =
So, if we can find a Ψ such that L(Ψ) = e^2 it, then L
= 1 + e^2 it.
Now, we assume that Ψ is of the form Ψ = v(t)e^2 it Ψ′^ = v′e^2 it^ + 2ive^2 it Ψ′′^ = v′′e^2 it^ + 4iv′e^2 it^ − 4 ve^2 it, where v is a polynomial. Now, if we plug this guess into the (modified) DE, we have Ψ′′^ e^2 it(v′′^ + 4iv′^ − 4 v) −2Ψ′^ e^2 it(− 2 v′^ − 4 iv) +5Ψ e^2 it(5v) e^2 it(1) = e^2 it(v′′^ + v′(4i − 2) + v(1 − 4 i)). Now, we see that if v is a mth^ degree polynomial, then v′′^ + v′(4i − 2) + v(1 − 4 i) is also mth^ degree. We need this to be a zero-th degree polyonmial, i.e., m = 0, and so v is of the form v = a,
whereby 0 + 0 + a(1 − 4 i) = 1 ⇒ a = 1 1 − 4 i = 1 + 4i 1 + 16 = 1 + 4i 17
Thus,
Ψ = 1 + 4i 17 e^2 it
=
cos(2t) − 4 sin(2t)
sin(2t) + 4 cos(2t)
Thus, L
= 1 + e^2 it^ = 1 + cos(2t) + i sin(2t).
Therefore, by equating real parts (and using linear operator theory)
L
= 1 + cos(2t),
and so ψ(t) =
cos(2t) − 4 sin(2t)
does the trick. (i.e., L(ψ) = 1 + cos(2t) = 2 cos^2 (2t).)
- y′′^ − 2 y′^ + 5y = 2(cos^2 t)et
Solution: First, notice that by the half-angle identity,
2 cos^2 tet^ = 2
1 + cos(2t) 2
et^ = et^ + et^ cos(2t).
So, let us study two related problems: L(y) = y′′^ − 2 y′^ + 5y = et, (2) and L(y) = y′′^ − 2 y′^ + 5y = ete^2 it^ = e(1+2i)t^ = eαt, (3) where α = 1 + 2i.
Suppose we have ψ 1 that satisfies (2) and ψ 2 that satisfies (3). Then L(ψ 1 + ψ 2 ) = L(ψ 1 ) + L(ψ 2 ) = et^ + eαt^ = et
1 + e^2 it
= et
1 + cos(2t) + i sin(2t)
If so, then the real part of ψ 1 +ψ 2 satisfies the real part of the righthand side, namely et
1+cos(2t)
2 cos^2 (t)et.
Let us find such a ψ 1. Assume that ψ 1 is of the form ψ 1 = etp(t) ψ 1 ′ = etp′^ + etp ψ 1 ′′ = etp′′^ + et 2 p′^ + etp, where p is a polynomial of degree m. Then ψ′′ 1 et(p′′^ + 2p′^ + p) − 2 ψ′ 1 et(− 2 p′^ − 2 p) +5ψ 1 et(+5p) et(1) = et(p′′^ + 4p)
- y′′^ − 3 y′^ + 2y = et^ + e^2 t
Hint: I’d do these two problems separately:
L(y) = et
and L(y) = e^2 t. Find ψ 1 for the first and ψ 2 for the second, and add the results.
- y′′^ + y = cos t cos 2t
Hint: This one is trickier. Consider the following:
cos(t) cos(2t) =
eit^ + e−it 2
e^2 it^ + e−^2 it 2
e^3 it^ + eit^ + e−it^ + e−^3 it
(2 cos(3t) + 2 cos(t))
=
cos(3t) +
cos(t).
So, I’d consider two related problems: L(y) =
e^3 it and L(y) =
2 e
it.
Find ψ 1 and ψ 2 that solve each, respectively. Add their results, and take the real part.
