Mathematics 4410
Homework 2
Answers
1. Find the value of A for which the solution of the initial value problem
y 0 − y = 1 + 3 sin t,
y(0) = A
remains nite as t → ∞.
Answer : We have e−t as an integrating factor, and multiplication by the integrating factor yields:
e−t y 0 − e−t y = e−t + 3e−t sin t
[e−t y] 0 = e−t + 3e−t sin t
Z
e−t y = (e−t + 3e−t sin t) dt
−t
= −e−t + 3 e 2 (− sin t − cos t) + C
y = −1 − 32 (sin t + cos t) + Cet
This solution will be bounded provided that C = 0. When C = 0, we see that y(0) = − 52 .
2. A home buyer can aord a mortgage payment of $1500/month. Suppose that the interest rate is 5.25%,
interest is compounded continuously, and that the payments are made continuously.
(a ) Determine the maximum amount of money that the buyer can aord to borrow on a 30-year mortgage.
(b ) Determine the total interest paid during that mortgage.
(c ) Repeat these calculations for a 20-year mortgage.
Answer : Suppose that the buyer borrows M dollars. We work in terms of years, so that the interest on M
dollars will be 0.0525M, and the payments will be $18000/yr. Therefore, the dierential equation becomes
M 0 = 0.0525M − 18000. Our initial condition is M(0) = C, where C is unknown, but we have M(30) = 0.
We rewrite the dierential equation as M 0 = 0.0525(M − 342857.1429), so that M 0 /(M − 342857.1429) =
0.0525. Integration results in log |M − 342857.1429| = 0.0525t + C, or |M − 342857.1429| = De0 0525t ,
and nally M = 342857.1429 + De0 0525t . We need to solve M(30) = 0, or −342857.1429 = De1 5750 ,
so that D = −70974.0181. Our equation becomes M = 342857.1429 − 70974.0181e0 0525t . The value of
M(0) = 271883.1248, which is the amount that can be borrowed.
The total amount of money paid over 30 years is $540000, so the amount of interest is 268116.8752.
Suppose instead that the term is 20 years. Then we set M(20) = 0, or −342857.1429 = De1 0500 ,
so that D = −119978.6568. The equation becomes M = 342857.1429 − 119978.6568e0 525t . We see that
M(0) = 222878.4860. The total amount of money paid over 20 years is 360000, so that the interest payments
total 137121.5140.
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3. Variation of Parameters is another method for solving rst-order linear dierential equations. Write
such an equation as
y 0 + p(t)y = g(t).
(1)
(a ) Suppose rst that g(t) = 0. Show that the solution is
Z
y = A exp − p(t) dt
for any constant A.
(b ) Now suppose that g(t) is not identically 0. Suppose that we try to nd a solution of (1) in the form
Z
y = A(t) exp − p(t) dt
(2)
where now A(t) is a function rather than a constant. Substitute expression (2) into the original
dierential equation (1) and conclude that A(t) must satisfy
A (t) = g(t) exp
0
Z
p(t) dt .
(3)
Using the expression for A 0 (t) in (3), it is possible to compute A(t). (Don't forget the constant of
integration!) It is then simple to substitute the function A(t) back into (2) to compute y.
(c ) Use this method to solve
y 0 − 2y = t2 e2t .
Answer :
0
0
We
R y /y =
−p(t). Integration yields
R rst set g(t) = 0, and solve y + py = 0. This becomes
log |y| = − p(t) dt, and the usual exponentiation
yields
y
=
A
exp
−
p(t)
dt
.
R
Now suppose that we set y = A(t) exp − p(t) dt . We have
Z
Z
y 0 = A 0 (t) exp − p(t) dt − A(t)p(t) exp − p(t) dt
Z
Z
y 0 + p(t)y = A 0 (t) exp − p(t) dt − A(t)p(t) exp − p(t) dt + p(t)y
Z
Z
Z
= A 0 (t) exp − p(t) dt − A(t)p(t) exp − p(t) dt + p(t)A(t) exp − p(t) dt
Z
= A 0 (t) exp − p(t) dt
We need to solve y 0 + py = g(t), so we are led to
Z
g(t) = A 0 (t) exp − p(t) dt
and so
A (t) = g(t) exp
0
Z
p(t) dt
Now, we applyRthis method
to solve y 0 − 2y = t2 e2t . We have p = −2 and g(t) = t2 e2t . Therefore,
2 2t
2 2t −2t
A (t) = t e exp −2 dt = t e e
= t2 , A(t) = t3 /3 + C, and the solution is
0
y=
Z
e2t t3
t3
+ C exp −(−2) dt =
+ Ce2t .
3
3
4. Solve the initial value problem
y 0 = 2(1 + x)(1 + y2 ),
y(0) = 0.
If possible, write y explicitly as a function of x.
Answer : This equation is separable. We have
dy
= 2(1 + x)(1 + y2 )
dx
Z
Z
dy
=
(2 + 2x) dx
1 + y2
arctan y = 2x + x2 + C
y = tan(2x + x2 + C)
Substitute x = 0 and y = 0 and we have:
y = tan(2x + x2 )
5. Suppose that, as usual, we write a rst-order dierential equation as y 0 = f(x, y). If f(x, y) can be written
as a function of the ratio y/x, then the function f(x, y) is called homogeneous. (We will unfortunately
use the same word \homogeneous" to mean something completely dierent in a few weeks. There are,
alas, not enough words in the English language to describe all useful mathematical concepts, so words are
2
2
, because we can
sometimes used more than once.) An example of a homogeneous function f(x, y) is xx2 −y
+y2
1−(y/x)2
x2 −y2
write x2 +y2 = 1+(y/x)2 .
Consider the dierential equation
dy
y
= + ey/x
dx
x
(4)
dv
(a ) Let v = y/x, so that y = xv. Compute dy
dx in terms of x, v, and dx .
dv
(b ) Substitute into (4) to derive a dierential equation for dx .
(c ) The dierential equation from the previous step is separable. Solve it, and then substitute v = y/x to
produce a solution of the original dierential equation (4) in terms of x and y. Your solution will be
an equation relating x and y; if possible, solve for y in terms of x.
The substitution v = y/x will always turn a homogeneous equation into a separable one.
dv
Answer : If y = xv, then dy
dx = v + x dx . Therefore, we have
v+
Z
dy
dx
dv
x dx
dv
x dx
=
y
x
+ ey/x
= v + ev
= ev
Z
dx
−v
e dv =
x
−e−v = log |x| + C
e−v = D − log |x|
−v = log(D − log |x|)
v = − log(D − log |x|)
y
x
= − log(D − log |x|)
y = −x log(D − log |x|)
6. Solve y 0 + y2 sin t = 0.
Answer : This equation is separable. We get
y 0 + y2 sin t = 0
y 0 = −y2 sin t
Z
y−2 y 0 = − sin t
Z
y−2 dy = − sin t dt
−y−1 = cos t + C
y=
1
D−cos t
.
7. Suppose that a sum of money S is invested at an annual rate of return r compounded continuously, so
that S 0 = rS.
(a ) Find the time T that it takes for the amount of money to double as a function of r.
(b ) If r = 3.25%, compute T .
(c ) If you would like your money to double in 10 years, what is r?
Answer :
We know that the solution of S 0 = rS is S = Cert . We have S(0) = C, and we need to solve
S(T ) = 2S(0) = 2C. Therefore, 2 = erT , and so T = (log 2)/r. Therefore, if r = 3.25%, we have T ≈ 21.3276
years. Conversely, if we set T = 10, we have r = (log 2)/T ≈ 0.0693 ≈ 6.93%.
8. Suppose that a young person with no initial savings decides to save k dollars per year at an annual
rate of return r. Assume that the investments are made continuously and that the return is compounded
continuously. (These unrealistic assumptions allow us to model the situation using a dierential equation.)
(a ) Determine S(t), the amount of money accumulated at time t.
(b ) Suppose that r = 4%, and our young person wishes to have accumulated $1 million after 40 years.
What is k?
(c ) Suppose that k is $2000/yr, and our young person wishes to have accumulated $1 million after 40 years.
What is r?
Answer : The dierential equation is S 0 = rS + k. Rewrite this as S 0 = r(S + k/r), and for simplicity
set k/r = a. We have S 0 /(S + a) = r, and we can as usual integrate: log |S + a| = rt + C, and so
S = −a + Dert = − kr + Drt . Our assumption that S(0) = 0 tells us that D = kr , so the equation becomes
S = kr (ert − 1).
Suppose rst that r = 0.04, and S(40) = 106 . We have 106 = 0 k04 (e1 6 − 1) ≈ 0 k04 (3.9530) ≈ 98.8258k,
and therefore k ≈ 10118.8140.
The third part of the problem is trickier, because r appears in two dierent places in the equation.
40r
40r
− 1), or 500 = e r −1 . Newton's method (or numerical experimentation) yields
We have 106 = 2000
r (e
r ≈ 0.097733.
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9. Newton's law of cooling states that the temperature of an object changes at a rate proportional to the
dierence between its temperature and that of its surroundings. Suppose that the temperature of a cup of
tea obeys this law. Suppose that the tea has a temperature of 95◦ C when poured, and has cooled to 91◦ C
after one minute, and that the temperature of the room is 21◦ C. How long does it take for the tea to cool
to 80◦ C?
Answer : Let T be the temperature as a function of time t measured in minutes. We have T 0 = k(T − 21).
We have as usual log |T − 21| = kt + C, and so T = 21 + Dekt . We have T (0) = 95 and T (1) = 91. The
rst equation gives 95 = 21 + D, and the second gives 91 = 21 + Dek . The rst equation says that D = 74,
and then the second says that 70 = 74ek , meaning that k ≈ −0.0556. The solution is T = 21 + 74e−0 0556t .
Substitute T = 80 and solve for t, and we get t ≈ 4.0764 min.
.