Homework: Review Problems
1. 2.3.5, 2.3.6 (these are to verify the trig integrals, given the trig identities).
Here is a restatement of the problem. Given the following trig identities:
1
cos(A) cos(B) = (cos(A − B) + cos(A + B))
2
1
sin(A) sin(B) = (cos(A − B) − cos(A + B))
2
cos(A + B) = cos(A) cos(B) − sin(A) sin(B)
sin(A + B) = sin(A) cos(B) + cos(A) sin(B)
Fun Side Remark: We can prove the last two using Euler’s Formula, since
cos(A + B) + i sin(A + B) = ei(A+B)
But, getting back to the problem, go ahead and integrate- Nothing tricky, just be sure
you consider the different cases.
2. Define the hyperbolic sine and cosine (sometimes called “sinch” and cosh”, or some
people say “sin-h and cos-h”)
cosh(x) =
ex + e−x
2
sinh(x) =
ex − e−x
2
Graphically, cosh(x) is parabola-like and is even, while sinh(x) is cubic-like and odd.
(a) Give derivative and integral formulas for the hyperbolic sine and cosine.
(b) Show that sinh(kx) and cosh(kx) form a fundamental set of solutions to y 00 = k 2 y,
at x = 0.
(c) Given the ODE: y 00 = 9y, show that not only do cosh(3x) and sinh(3x) solve the
DE, but so do translations: cosh(3(x − L)) and sinh(3(x − L)) for any given L.
SOLUTIONS: I think these are straightforward, but let me know if you have any difficulty with them.
3. Review: Euler’s Equation is of the form
x2 y 00 + αxy 0 + βy = 0
We use an ansatz of y = xr and we get the characteristic equation:
r(r − 1) + αr + β = 0
Solve it using the quadratic formula, and we have three types of solutions based on the
discriminant. This can be summarized below.
• Two real roots, r1 , r2 : y = C1 xr1 + C2 xr2
• One root: y(x) = xr (C1 + C2 ln(x))
1
• Complex roots, r = α ± βi the solutions are the real and imaginary parts of xα+βi ,
which are:
y(x) = xα (C1 cos(β ln(x)) + C2 sin(β ln(x))
Solve the following Euler equations:
(a) x2 y 00 + 3xy 0 + 5y = 0
SOLUTION: With the ansatz y = xr , we substitute y 0 = rxr−1 and y 00 = r(r −
1)xr−2 to get:
r(r − 1)xr + 3rxr + 5xr = 0
⇒
r(r − 1) + 3r + 5 = 0
⇒
r2 + 2r + 5 = 0
(r + 1)2 + 4 = 0 r = −1 ± 2i
The solutions are the real part and imaginary part of
x−1+2i = x−1 x2i = x−1 e2i ln(x) = x−1 (cos(2 ln(x)) + i sin(2 ln(x))
Therefore, the solution to the differential equation is:
y=
1
(C1 cos(2 ln(x)) + C2 sin(2 ln(x)))
x
(NOTE: We’ll assume that x > 0 to avoid problems with existence and discontinuities)
(b) x2 y 00 + 4xy 0 + 2y = 0
SOLUTION: Using the same ansatz as before, the characteristic equation becomes:
r(r − 1) + 4r + 2 = 0
r2 + 3r + 2 = 0
⇒
⇒
(r + 2)(r + 1) = 0
We have two distinct real roots:
y(x) = C1 x−2 + C2 x−1
(c) x2 y 00 − 3xy 0 + 4y = 0
SOLUTION: Using the same ansatz, the characteristic equation becomes:
r(r − 1) − 3r + 4 = 0
r2 − 4r + 4 = 0
⇒
so r = 2 is a double root.
y(x) = x2 (C1 + C2 ln(x))
2
⇒
(r − 2)2 = 0