Math 215
Spring 2010
Assignment 1
§2.1: 13, 15, 30, §2.2: 3, 10, 21, §2.3: 2, 8, 16, §2.4: 1, 15, 22(a,b)
• (§2.1 #13) Solve y 0 − y = 2te2t ,
Answer.
y(0) = 1.
µ(t) = e−t so that (e−t y)0 = 2tet and thus
Z
Z
−t
t
t
e y(t) = 2te dt = 2te − 2 et dt = 2(t − 1)et + c.
Setting t = 0 and y = 1 we get 1 = −2 + c, or c = 3. Hence y(t) = 2(t − 1)e2t + 3et .
• (§2.1 #15) Solve ty 0 + 2y = t2 − t + 1,
y(1) = 12 ,
t > 0.
0
Answer. Rewrite
with p(t) = 2/t and g(t) = t − 1 + 1/t. Take
R 2dt as 2y + p(t)y =2 g(t)
µ(t) = exp( t ) = t so that (t y)0 = t3 − t2 + t. Integrating and dividing by t2
gives y = t2 /4 − t/3 + 1/2 + c/t2 . Setting t = 1 and y = 1/2 we have c = 1/12. Thus
1
y = t2 /4 − t/3 + 1/2 + 12t
2.
• (§2.1 #30) Find the value of y0 for which the solution of the initial value problem
y 0 − y = 1 + 3 sin t,
y(0) = y0
remains finite as t → ∞ (i.e., uniformly bounded by some constant for all t).
Answer.
(e−t y)0 = e−t + 3e−t sin t so e−t y(t) = −e−t − 23 e−t (sin t + cos t) + c, or
3
y(t) = −1 − (sin t + cos t) + cet .
2
Setting t = 0 we get y0 = −1 − 3/2 + c. If y(t) is to remain bounded as t → ∞, we
must have c = 0 and thus y0 = −5/2.
• (§2.2 #3) Solve y 0 + y 2 sin x = 0.
Answer.
If y(x0 ) = 0 for some x0 , then y(x) = 0 for all x. If y 6= 0, the DE may
be written as y −2 dy = − sin x dx. Integrating both sides, we get −y −1 = cos x + c, or
y = −1/(cos x + c).
Note: For any given initial data y(x0 ) = y0 6= 0, we have cos x + c 6= 0 for x near x0 .
The interval of existence is finite if |c| ≤ 1.
• (§2.2 #10) Consider
y 0 = (1 − 2x)/y,
(a) Find the solution in explicit form.
(b) Plot the graph of the solution.
y(1) = −2.
(c) Determine the interval in which the solution is defined.
Answer.
(a) Rewrite as ydy = (1 − 2x)dx. √Integrating we get y 2 /2 = x − x2 + c.
Using y(1) = −2 we get c = 2. Thus y(x) = − 4 + 2x − 2x2 , where the negative sign
is due to y(1) < 0.
(b)
(c) The solution is defined if 4+2x−2x2 > 0. Thus the interval of existence is (−1, 2).
• (§2.2 #21) Solve
y 0 = (1 + 3x2 )/(3y 2 − 6y),
y(0) = 1
and determine the interval in which the solution is valid.
Hint: look for points where the integral curve has a vertical tangent.
Answer. We have (3y 2 − 6y)dy = (1 + 3x2 )dx so that y 3 − 3y 2 = x + x3 + c. Using
y(0) = 1 we get c = −2. From the DE, the integral curve has a vertical tangent when
3y 2 − 6y = 0, i.e., y = 0 or y = 2. For y = 0 we have x3 + x − 2 = 0. Its only real
zero is x = 1. For y = 2 we have x3 + x + 2 = 0. Its only real zero is x = −1. Thus
the interval of existence is x ∈ (−1, 1).
• (§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 wellstirred mixture leaves the tank at the same rate. Find an expression in terms of γ for
the amount of salt in the tank at any time t. Also find the limiting amount of salt in
the tank as t → ∞.
Answer. Let S(t) be the amount of salt at time t. Then S(0) = 0. We have rate in
= 2γ and rate out = 2(S/120), thus
dS/dt = 2γ − S/60,
S(0) = 0.
This is a linear equation and we can take integrating factor et/60 , resulting in (et/60 S)0 =
2γet/60 . Integrating and dividing by et/60 , we get S(t) = 120γ + ce−t/60 . The condition S(0) = 0 gives c = −120γ, so S(t) = 120γ(1 − e−t/60 ) and hence S(t) → 120γ as
t → ∞.
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• (§2.3 #8) A young person with no initial capital invests k dollars per year at an annual
rate of return r. Assume that investments are made continuously and that the return
is compounded continuously.
(a) Determine the sum S(t) accumulated at any time t.
(b) If r = 7.5%, determine k so that $1 million will be available for retirement in 40
years.
(c) If k = $2000/year, determine the return rate r that must be obtained to have $1
million available in 40 years.
Answer.
(a) It satisfies S(0) = 0 and dS/dt = rS + k. Thus (e−rt S)0 = ke−rt ,
−rt
e S(t) = − kr e−rt + c. The condition S(0) = 0 gives c = kr . Thus S(t) = kr (ert − 1).
k
3
0.075 (e − 1). Thus k = 3929.68 ($/year).
40r − 1). Thus 500r = e40r − 1 and
106 = 2000
r (e
(b) r = .075 and S(40) = 106 gives 106 =
(c) k = 2000 and S(40) = 106 gives
r =9.77%.
(The last rate can be obtained by trial and error, or by Newton’s method.)
• (§2.3 #16) Newton’s law of cooling states that the temperature of an object changes
at a rate proportional to the difference between its temperature and that of its surroundings. Suppose that the temperature of a cup of coffee obeys Newton’s law of
cooling. If the coffee has a temperature of 200 ◦ F when freshly poured, and 1 min
later has cooled to 190 ◦ F in a room at 70 ◦ F, determine when the coffee reaches a
temperature of 150 ◦ F.
Answer.
If T (t) is the temperature of the coffee at any time t, then
dT /dt = −k(T − 70),
T (0) = 200,
T (1) = 190.
Taking integrating factor ekt we have (ekt T )0 = 70kekt . Integrating and dividing by
ekt , we get T (t) = 70 + ce−kt . The two initial conditions give
200 = 70 + c,
190 = 70 + ce−k .
Thus c = 130, e−k = 120/c = 12/13, and T (t) = 70+130(12/13)t . Solving T (t) = 150,
ln(8/13)
we get (12/13)t = 8/13 and thus t =
= 6.07 (min).
ln(12/13)
• (§2.4 #1) Determine (without solving the problem) the interval of validity of the
following problem
(t − 3)y 0 + (ln t)y = 2t, y(1) = 2.
ln t
2t
Answer.
Rewrite the DE as y 0 + t−3
y = t−3
, which is a linear equation. The
solution exists as long as the coefficients remain finite, that is, t > 0 and t 6= 3. This
interval contains t = 1, and is thus (0, 3).
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• (§2.4 #15) Solve the problem
y 0 + y 3 = 0,
y(0) = y0 ,
and determine how the interval in which the solution exists depends on the initial
value y0 .
Answer.
The equation is separable, with dy/y 3 = −dt. Integrating both sides and
y0
invoking the initial condition, y −2 = t + y0−2 and thus y(t) = p
. In deciding
1 + 2ty02
the sign when taking square root, we have used that y(t) has the same sign as y0 .
Solutions exist as long as 1 + 2ty02 > 0.
If y0 6= 0, solutions exist for t > − 2y12 .
0
If y0 = 0, the solution y(t) = 0 exists for all t.
• (§2.4 #22(a,b)) (a) Verify that both y1 (t) = 1 − t and y2 (t) = −t2 /4 are solutions of
the initial value problem
y0 =
−t + (t2 + 4y)1/2
,
2
y(2) = −1.
Where are these solutions valid?
(b) Explain why the existence of two solutions of the given problem does not contradict
the uniqueness part of Theorem 2.4.2.
Answer.
side)
(a) For y = y1 (t) = 1 − t, y(2) = −1, y 0 = −1 while (RHS = right hand
1
1
RHS(DE) = (−t + (t2 + 4 − 4t)1/2 ) = (−t + |t − 2|)
2
2
which is −1 if t ≥ 2 and 1 − t if t < 2. Thus y1 satisfies the DE for t ≥ 2.
For y = y2 (t) = −t2 /4, y(2) = −1, y 0 = −t/2 while
1
RHS(DE) = (−t + (t2 − t2 )1/2 ) = −t/2
2
which is y 0 . Thus y2 satisfies the DE for all t ∈ R.
(b) By Theorem 2.4.2 we are guaranteed a unique solution only where f (t, y) =
1
2
1/2 ) and f (t, y) = (t2 + 4y)−1/2 are continuous. In this case the initial
y
2 (−t + (t + 4y)
point (2, −1) lies in the region t2 + 4y ≤ 0 so fy is not continuous and hence the
theorem is not applicable and there is no contradiction.
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