MATH 246 - Quiz 2 Solutions
Instructions. Solve the two problems below on this sheet of paper. You may use the reverse
side. Show all work.
1. (5 points) Solve the following initial value problem by separation of variables:
dy
t(1 + y 2 )
=
,
dt
1 + t2
y(0) = 0.
Express the (implicit) solution that you obtain explicitly in the form y = ϕ(t), √
i.e., solve for
y explicitly as a function of t. Now explain why that solution blows up at t = eπ − 1.
We begin by noting that this differential equation is first-order, nonlinear, and separable.
Thus, we begin by separating variables, placing all terms with y on the left-hand side and
all terms with t on the right-hand side.
dy
t
=
dt
2
1+y
1 + t2
Next, we integrate both sides.
Z
dy
=
1 + y2
Z
t
dt
1 + t2
(1)
1
For the left-hand side, we note that the antiderivative of 1+y
2 is arctan(y). For the right
2
hand side, we make the substitution u = 1 + t , giving us that du = 2t dt and thus that
du
dt = 2dt
. Making this substitution, we have the integral
Z
11
1
1
du = log |u| = log |1 + t2 |
2u
2
2
Substituting these evaluated integrals back into (1) and remembering to include a constant
of integration on the right-hand side, we obtain the general implicit solution
arctan(y) =
1
log(1 + t2 ) + C
2
(2)
where we have dropped the absolute value because 1 + t2 > 0 for all values of t.
Next, we solve for C by recalling the initial condition y(0) = 0. Thus, we substitute 0 in for
y and 0 in for t, arriving at
1
arctan(0) = log(1) + C
(3)
2
Both arctan(0) and log(1) evaluate to 0, and thus we know that C = 0 to maintain equality.
Now we may solve for y explicitly as a function of t,
1
2
y = tan
log(1 + t )
(4)
2
√
eπ − 1, we substitute this value into y(t), giving us
√
√
1
2
π
π
y( e − 1) = tan
log(1 + ( e − 1) )
(5)
2
π = tan
(6)
2
To determine why y blows up at t =
Note that tangent has a vertical asymptote at π/2, and thus we expect the solution to blow
up there.
2
2. (5 points) Joe Smith inherits $100,000. He places it in a new account that pays interest
continuously at an annual rate of 5%. Joe estimates that (in addition to Social Security
and his pension), he needs the account to be worth $300,000 on the day that he retires.
How many more years must Joe continue to work? Alternatively, suppose that in addition
to the initial deposit, Joe puts $10,000 per year into the account on a continuous basis.
Now how long must he work? Hints: ln 3 ∼ 1.1; ln(5/3) ∼ .5; and the solution of the IVP
S 0 = rS + k, S(0) = S0 is S = (S0 + kr )ert − kr .
We are given the differential equation modeling the amount S in an account with continuous
interest and continuous investment, namely
S 0 = rS + k, S(0) = S0
as well as its solution
S=
k
S0 +
r
ert −
k
r
(7)
(8)
Thus, it suffices to identify the values for the parameters k, r, and S0 . The parameter k
is the amount invested every year, $0 in the case of the first scenario, the parameter r is
the interest rate, 5% = 0.05, and the parameter S0 is the initial amount of money invested,
$100,000. Substituting in these values, we find
S = 1 × 105 e0.05t
We wish to find the time tf in years for which S(tf ) = 3 × 105 . Substituting in these values
and performing algebraic manipulations to solve for tf , we find
3 × 105 = 1 × 105 e0.05tf
3 = e0.05tf
log(3) = 0.05tf
tf = 20 log(3) ≈ 20 · 1.1 = 22 years
We now wish to consider the case where Joe continuously invests $10,000 continuously every
year. In this case, k = 1 × 104 . Substituting this value for k into (8), we have
1 × 104 0.05t 1 × 104
5
S = 1 × 10 +
e
−
0.05
0.05
5 0.05t
5
= 3 × 10 )e
− 2 × 10
Again, we wish to find the time tf for which S(tf ) = 3 × 105 . Making this substitution and
solving for tf , we find
3 × 105 = 3 × 105 )e0.05tf − 2 × 105
5 = 3e0.05tf
3
5
log
= 0.05tf
3
5
tf = 20 log
≈ 20 · 0.5 = 10 years
3
Thus, we see that with continuous investment, it will take less than half the amount of time
to reach the target goal of $300,000.
4