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Ordinary Differential Equations, Spring 2015
Solutions to Homework 2
Maximal grade for HW3: 100 points
Section 2.2 Solve the initial value problem, sketch the graph of the
solution and determine the interval of its definition.
12. dr/dθ = r2 /θ, r(1) = 2.
Solution: This is a separable differential equation:
Z
Z
dr
dθ
1
=
⇒ − = ln |θ| + C.
2
r
θ
r
For θ = 1 we have −1/r(1) = −1/2 = C, so
r=
−1
−1
=
.
ln |θ| + C
ln |θ| − 1/2
1/2
The
√ solution is not defined for θ = 0 and for ln |θ| = 1/2 ⇔ θ = ±e =
e. Therefore the natural interval of its definition containing θ = 1 is
± √
(0, e).
3
2
1
−1
1
2
3
−1
√
16. y 0 = x(x2 + 1)/4y 3 , y(0) = −1/ 2.
Solution: This is a separable differential equation:
Z
Z
x 4 x2
3
4y dy = x(x2 + 1)dx ⇒ y 4 =
+
+ C.
4
2
1
At x = 0 we have C = 1/4, so
y4 =
1
1 √
x4 x2 1
+
+ = (x2 + 1)2 , y(x) = − √ x2 + 1.
4
2
4
4
2
Note that we choose the negative branch of the square root since y(0) < 0.
The solution is defined everywhere.
−3
−2
−1
1
2
3
−2
−4
23. Solve the initial value problem y 0 = 2y 2 + xy 2 , y(0) = 1 and determine
where the solution attains its minimum value.
Solution: This is a separable differential equation:
Z
Z
dy
1
x2
dy
2
= (2 + x)y ,
=
(2
+
x)dx
⇒
−
=
2x
+
+ C.
dx
y2
y
2
Since y(0) = 1 we have C = −1, so
y=
x2
2
−1
.
+ 2x − 1
Let us find the interval of definition of y(x). We have
√
x2
+ 2x − 1 == 0 ⇔ x2 + 4x − 2 = 0 ⇔ x = −2 ± 6.
2
√
√
Therefore the solution is defined on (−2 − 6, −2 + 6). The minimal value
of y(x) corresponds to the maximal value of x2 + 4x − 2 which is attained at
x = −2. Note that this point belongs to the interval of definition.
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Section 2.3 2. A tank initially contains 120 L of pure water, A mixture
containing a concentration of γ g/L of salt enters the tank at a rate of 2
L/min, and the well-stirred mixture leaves the tank at the same rate. Find
an expression in terms of γ of the amount of salt in the tank at any time t.
Also find the limiting amount of salt in the tank as t → ∞.
Solution: Let y(t) denote the mass of salt in grams at time t, then the
concentration in the mixture leaving the tank (and the concentration in the
tank) equals y(t)/120 g/L. Therefore the differential equation for y(t) has
the form:
120γ − y(t)
.
y 0 (t) = 2γ − 2 · y(t)/120 =
60
This is a separable differential equation:
Z
Z
dt
dy
=
, − ln |120γ − y| = t/60 + C, ln |120γ − y| = −t/60 − C,
120γ − y
60
and for A = ±e−C we have
120γ − y = Ae−t/60 .
Since y(0) = 0 (initially the tank contained only pure water and no salt), we
get A = 120γ, so
y(t) = 120γ − 120γe−t/60 .
At t → +∞ the mass tends to y(t) → 120γ.
12. A recent college graduate borrows 150000 at an interest rate of 6%
to purchase a condominium. Anticipating steady salary increases, the buyer
expects to make payments at a monthly rate of 800 + 10t, where t is the
number of months since the loan was made.
(a) Assuming that this payment schedule can be maintained, when will
the loan be fully paid?
(b) Assuming the same payment schedule, how large a loan could be paid
off in exactly 20 years?
Solution: Let y(t) be the loan amount after t months. The monthly
interest rate is r = 0.06/12 = 0.005, and the differential equation for y(t)
has the form:
y 0 (t) = 0.005y(t) − 800 − 10t ⇔ y 0 (t) − 0.005y(t) = −800 − 10t.
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This is a linear differential equation with an integrating factor µ(t) = e−0.005t ,
so we get
(e−0.005t y(t))0 = e−0.005t y 0 (t) − 0.005e−0.005t y(t) = −800e−0.005t − 10te−0.005t .
Therefore
Z
800 −0.005t
10
dt =
−800e
− 10te
e
+
td(e−0.005t ) =
e
y(t) =
0.005
0.005
Z
10
10
−0.005t
−0.005t
160000e
+
te
−
e−0.005 dt =
0.005
0.005
2000 −0.005
e
+C = 2000te−0.005t +560000e−0.005t +C,
160000e−0.005t +2000te−0.005t +
0.005
so
y(t) = 2000t + 560000 + Ce0.005t .
−0.005t
Z
−0.005t
−0.005t
(a) If y(0) = 150000 then C = −410000 and
y(t) = 2000t + 560000 − 410000e0.005t .
One can check that y(t) = 0 for t ≈ 146.5 months.
(b) If the loan is repayed in exactly 20 years (=240 months), we have
y(240) = 0, so
0 = y(240) = 2000 · 240 + 560000 + Ce0.005·240 = 1040000 + Ce1.2 ,
and C = −e−1.2 · 1040000 ≈ −313242. The initial loan amount equals
y(0) = 560000 + C = 560000 − 313242 = 246758.
Section 2.5 7. Solve the differential equation dy/dt = k(1−y)2 for k > 0
and y(0) = y0 .
Solution: This is a separable differential equation:
Z
Z
1
1
dy
= kdt,
= kt + C, 1 − y =
,
2
(1 − y)
1−y
kt + C
so
y(t) = 1 −
4
1
.
kt + C
Since y 0 (t) ≥ 0 for all t, all solutions are increasing. At t = 0 we get
C=
1
,
1 − y0
so C is positive for y0 < 1 and negative for y0 > 1. The solution is undefined
for t = −C/k, so the interval of definition of y(t) for the initial value problem
y(0) = y0 equals (−∞, −C/k) = (−∞, k(y01−1) ) for y0 > 1 and ( k(y01−1) , +∞)
for y0 < 1. In the first case the solution goes to infinity in finite time, in the
second case the solution approaches 1 at t → +∞. Finally, if y(0) = y0 = 1,
we get an equilibrium solution y(t) = 1.
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