MATH 101 HOMEWORK 10 – SOLUTIONS
1. A bank determines that the waiting time for a customer’s call to be answered by a
representative is modelled by an exponential density function whose expectation k depends
on the number of representatives they employ. If the bank wants 90% of the calls to be
answered within the first 7 minutes, what value of k should it aim for? (You may use a
calculator to finish the computations.)
The exponential density with expectation 1/k is ke−kx . We want 1/k to be small enough
so that
Z ∞
−7k
ke−kx dx = −e−kx |∞
≤ 0.1.
7 =e
7
This is possible if k ≥ (ln 10)/7 ≈ 0.329, which corresponds to the expectation 1/k ≥
7/ ln 10 ≈ 3.040.
2. Solve the differential equations:
(a) y 0 =
1+x
, y(1) = −4: this is a separable equation which we rewrite as
xy
ydy =
1
1+x
dx = ( + 1)dx.
x
x
Integrating both sides, we get y 2 /2 = ln |x| + x + C. To find C, we plugpin y = −4, x = 1:
16/2 = 0 + 1 + C, C = 8 − 1 = 7. Thus y 2 /2 = ln |x| + x + 7, y = − 2(ln |x| + x + 7).
(We took the negative square root because −4 is negative.)
(b) xy 0 − 3y = x4 cos x: we first rewrite this as a linear equation, i.e. y 0 − 3y/x = x3 cos x.
We find the integrating factor:
R
− (3/x)dx
r(x) = e
= e−3 ln |x| = |x|−3 .
We multiply the last equation by r(x). For x > 0, |x| = x, so we get
x−3 y 0 − 3x−4 = cos x, (x−3 y)0 = cos x, x−3 y = sin x + C, y = x3 sin x + Cx3 .
Note that this also works for x < 0, even though our formula for r(x) gives r(x) = −x−3
in this case! Why? (Look up the differential equation for r(x).)
3. End of term: everyone gets 5 marks for free!
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