MATH 54 − HINTS TO HOMEWORK 10
PEYAM TABRIZIAN
Here are a couple of hints to Homework 10. Make sure to attempt the
problems before you check out those hints. Enjoy!
S ECTION 4.1: I NTRODUCTION : T HE MASS - SPRING OSCILLATOR
4.1.1, 4.1.3. Just plug in those functions in the differential equation.
4.1.9. Plug in the functions with Ω = 2, and compare the coefficients,
just like you’d do for polynomials. For example, if you get A cos(2t) +
B sin(2t) = 6 cos(2t) + 8 sin(2t), then you’d say A = 6 and B = 8.
S ECTION 4.2: H OMOGENEOUS LINEAR EQUATIONS :
THE GENERAL
SOLUTION
4.2.29. Before you use the Wronskian, see if you can simplify your functions a little bit beforehand.
4.2.31. Use tan2 (x) + 1 = sec2 (x)
4.2.32. A set with the zero vector is always linearly dependent. Aren’t
functions just vectors?
4.2.35.
(a)
(b)
(c)
(d)
(hopefully) straightforward
Use sin2 + cos2 = 1
See how you can simplify your functions beforehand
x
−x
Remember that cosh(x) = e +e
2
4.2.37, 4.2.39. Use the technique we learned in Chapter 5 with factoring
out a polynomial of degree 3 (i.e. rational roots theorem and long division):
Rational roots theorem: If a polynomial p has a zero of the form r = ab ,
then a divides the constant term of p and b divides the leading coefficient of
p.
This helps you ‘guess’ a zero of p. Then use long division to factor out p.
Date: Tuesday, April 7th, 2015.
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PEYAM TABRIZIAN
S ECTION 4.3: AUXILIARY EQUATIONS WITH COMPLEX ROOTS
4.3.19, 4.3.29(a). Use the technique we learned in Chapter 5 with factoring
out a polynomial of degree 3 (i.e. rational roots theorem and long division),
see above.
4.3.30. You may use the fact that (f (t)+ig(t))0 = f 0 (t)+ig 0 (t), just expand
out the answer in (6) and differentiate.
4.3.37. For (a), it may be useful to let X = r2 , then the auxiliary equation
becomes X 2 + 2X + 1.
S ECTION 4.4: T HE METHOD OF UNDETERMINED COEFFICIENTS
4.4.3. Yes.
sin(x)
e4x
= e−4x sin(x).
4.4.5. Yes. 4x sin2 (x) + 4x cos2 (x) = 4x.
4.4.6. No
4.4.8. Yes
4.4.31. Guess:
yp (t) = (At3 +Bt2 +Ct+D)te−t cos(t)+(Et3 +F t2 +Gt+H)te−t sin(t)
(You have to add an extra t because e−t sin(t) coincides with the general
solution of the homogeneous equation)
4.4.33, 4.4.35. As usual, use the technique we learned in Chapter 5 with
factoring out a polynomial of degree 3 (i.e. rational roots theorem and long
division), see above.
S ECTION 4.5: T HE SUPERPOSITION PRINCIPLE
4.5.1. For example, for (b), by linearity, the solution is y(t) = 2y2 (t) −
3y1 (t)
MATH 54 − HINTS TO HOMEWORK 10
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4.5.21. Solution:
Homogeneous equation:
First of all, the auxiliary equation is r2 + 2r + 2 = 0, which gives:
r=
−2 ±
√
−2 ± 2i
4−8
=
= −1 ± i
2
2
which tells you the general solution of y 00 + 2y 0 + 2 = 0 is:
y0 (θ) = Ae−θ cos(θ) + Be−θ sin(θ)
Particular solution:
Notice that e−θ cos(θ) is already a solution of the homogeneous equation,
so we’ll have to guess:
yp (θ) = Aθe−θ cos(θ) + Bθe−θ sin(θ)
This gives us:
yp0 (θ) =Ae−θ cos(θ) − Aθe−θ cos(θ) − Aθe−θ sin(θ) + Be−θ sin(θ) − Bθe−θ sin(θ) + Bθe−θ cos(θ)
=(A − Aθ + Bθ)e−θ cos(θ) + (B − Bθ − Aθ)e−θ sin(θ)
And:
yp00 (θ) =(B − A)e−θ cos(θ) − (A − Aθ + Bθ)e−θ cos(θ) − (A − Aθ + Bθ)e−θ sin(θ)
+(−B − A)e−θ sin(θ) − (B − Bθ − Aθ)e−θ sin(θ) + (B − Bθ − Aθ)e−θ cos(θ)
−θ cos(θ)
− Bθ + B − Bθ − Aθ
Aθ)e
=(B − A − A + − B − A − B +
+ Aθ)e−θ sin(θ)
+(−A + Aθ − Bθ
Bθ
=(2B − 2A − 2Bθ)e−θ cos(θ) + (−2A − 2B + 2Aθ)e−θ sin(θ)
Plugging those formulas into our equation y 00 + 2y 0 + 2y = e−θ cos(θ),
we get:
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PEYAM TABRIZIAN
yp00 + 2yp0 + 2yp = e−θ cos(θ)
(2B − 2A − 2Bθ)e−θ cos(θ) + (−2A − 2B + 2Aθ)e−θ sin(θ) +
2 (A − Aθ + Bθ)e−θ cos(θ) + (B − Bθ − Aθ)e−θ sin(θ)
+2 Aθe−θ cos(θ) + Bθe−θ sin(θ) = e−θ cos(θ)
+
−θ
+
+ 2Aθ
−
−
2B − 2A
2Bθ
2A
2Aθ
2Bθ
e cos(θ)
+ 2B
2Bt
− 2Aθ
e−θ sin(θ)
+
+ −2A − 2B
2Aθ
2Bθ
− +
= e−θ cos(θ)
2Be−θ cos(θ) + (−2A)e−θ sin(θ) = 1e−θ cos(θ) + 0e−θ sin(θ)
Comparing the left-hand-side and the right-hand-side, we get 2B = 1 and
−2A = 0, so A = 0 and B = 12 , which tells us that a particular solution
is:
1
1
yp (θ) = 0θe−θ cos(θ) + θe−θ sin(θ) = θe−θ sin(θ)
2
2
General solution: And therefore the general solution to our differential
equation is:
1
y(θ) = y0 (θ) + yp (θ) = Ae−θ cos(θ) + Be−θ sin(θ) + θe−θ sin(θ)
2
4.5.27, 4.5.31, 4.5.36. It’s always better to plug each yp separately. For
example, in 4.5.27, first guess yp (t) = A cos(t) + B sin(t) and plug this
into the equation, and then guess yp (t) = A cos(2t) + B sin(2t) and plug
this into the equation.
4.5.39. As usual, use the technique we learned in Chapter 5 with factoring
out a polynomial of degree 3 (i.e. rational roots theorem and long division),
see above.
4.5.46. First find the general solution of the equation y(t) (in terms of λ),
then then solve for A and B using y(0) = 0 and y(π) = 1. At some point
you should find that y(π) = B sin(λπ) = 1, but if λ is an integer, you get
sin(λπ) = 0, and so y(π) = 0 = 1, which is impossible. But if λ is not an
1
integer, you get B = sin(λπ)
, which gives you a legit solution. Make sure
to treat the case λ = 1 separately! (since the root ±i will coincide with
the right-hand-side)