Math 266 Term Test 1
SOLUTIONS
QUESTIONS:
1. For the following give your answers as ”true” or ”false” only.
a) A first order ODE can be both linear and separable. True.
b) Every first order Bernoulli DE is a linear ODE. False.
c) A first order DE of the form M (x, y)dx+N (x, y)dy = 0 is exact
if Mx (x, y) = Ny (x, y). False.
d) The Euler’s Method always provides the actual solution of an
IVP. False.
e) A linear first order DE y 0 = f (x, y) with initial value has unique
solution if fy (x, y) is continuous in a region around the initial
value. True.
2. For the following IVP’s: (i) first determine if their DE’s are linear
or not, (ii) determine their order, and (iii) identify the type(s) of
first order ones.
a) y 0 + x(y 2 + y) = 0, y(2) = 1.
Answer: Nonlinear (y 2 ), 1st order, separable and Bernoulli.
, y(0) = 1.
b) y 0 − 4y = 48x
y2
Answer: Nonlinear (y −2 ), 1st order, Bernoulli.
c) (e−2x − 1)y 00 − y = 2x, y(0) = ln 2.
Answer: Linear, 2nd order.
d) xy 2 y 0 = x3 + y 3 , y(1) = 3.
Answer: Nonlinear (y 3 ), 1st order, (x,y)-homogeneous and
Bernoulli.
e) cos xy 0 = sin x − y sin x − 2 cos x, y(0) = 1.
Answer: Linear , 1st order, linear and exact.
3. Solve any two of the IVP’s in Question 2.
a) y 0 + x(y 2 + y) = 0, y(2) = 1. (Problem 12, p: 52.)
0
Solution: The DE can be rewritten as y2y+y = −x, a separable
R 0
R
R
2
1
DE. Thus y2y+y dx = − xdx iff y1 dy − y+1
dy = − x2 + C iff
2
y
ln | y+1
| = − x2 + C; hence, the general solution is
Since y(2) = 1, we have C =
IVP is
y
y+1
2
=
e2
;
2
y
y+1
x2
= Ce− 2 .
therefore, the solution of the
1 (2− x2 )
e
.
2
b) y 0 − 4y = 48x
, y(0) = 1. (Problem 11, p:69.)
y2
Solution: The DE is Bernoulli and can be rewritten as
y 2 y 0 − 4y 3 = 48x. Let v = y 3 , then v 0 = 3y 2 y 0 . Hence, multiplying both sides of the DE by 3, it takes the form v 0 −12v = 144x,
a linear
R DE. The, the associated integrating factor is
µ = e (−12) = e−12x ; and multiplying both sides of the last DE
1
2
with µ, we have [e−12x v]0 = 144xe−12x . Integrating, we get
e−12x v = −12xe−12x −e−12x +C; thus, y 3 = −(1+12x)+Ce12x is
the general solution. From the initial condition y(0) = 1, C =
2; hence, the solution of the IVP is y 3 = −(1 + 12x) + 2e12x .
c) (e−2x − 1)y 00 − y = 2x, y(0) = ln 2. This is not a 1st order IVP.
d) xy 2 y 0 = x3 + y 3 , y(1) = 3. (Problem 19, p:70.)
Solution: Dividing both sides of the DE, it takes the form
y 0 = xy +( xy )2 , an (x,y)-homogeneous DE. Letting v = xy , we have
v+xv 0 = y 0 . This leads to v+xv 0 = v12 +v; equivalently, v 2 v 0 = x1 ,
a separable DE. Integrating both sides, we have v 3 = 3 ln |x|+C;
and hence y 3 = x3 ln |x|3 + Cx3 is the general solution. Now,
from the initial condition y(1) = 3, C = 27; hence, the solution
of the IVP is y 3 = x3 ln |x|3 + 27x3 .
e) cos xy 0 = sin x − y sin x − 2 cos x, y(0) = 1. (Pr. 21, p: 80.)
(This DE is both linear and exact. It was solved as one of the
examples of exact DEs in the class. See your notes for that
solution. Here, another solution will be provided as a linear
DE.)
Solution: The DE can be rewritten as y 0 +(tan
x)y = tan x−2,
Rx
(tan u)du
a linear DE with integrating factor µ = e
= eln | sec x| =
sec x. Multiplying both sides of the DE by µ, we get
sin x
[(sec x)y]0 = cos
2 x − 2 sec x. Integrating both sides, we have
(sec x)y = − sec x−2 ln | sec x+tan x|+C. Therefore, the general
solution is y = −1−2 cos x ln | sec x+tan x|+C cos x. Now, from
the initial condition y(0) = 1, C = 2; hence, the solution of the
IVP is y = 2 cos x(1 − ln | sec x + tan x|) − 1.
4. Scientists at Los Alamos Nuclear Laboratory have discovered a new
radioactive element and named it Çömezium. It decays at the rate of
5% per hour, and the lab can produce the element continuously at a
rate of 2 grams/hour. If the original amount produced is 1 gr., derive
the formula that gives the amount of Çömezium at any time t > 0.
(Similar to Pr. 18 and 16 on pp: 138-139.)
Solution: The decay rate is r = −0.05, and 2 grams of Çömezium
per unit time (hr.) is added continuously. Hence, the rate of change
in the amount of Çömezium is P 0 (t) = −(0.05)P (t) + 2, which leads
to P 0 (t) + (0.05)P (t) = 2, a linear 1st order DE with initial condition
P (0) =
1 gr. The associated integrating factor is
R
(0.05)dt
µ=e
= e(0.05)t , from which we get [µP ]0 = 2e(0.05)t . Integrating,
e(0.05)t P (t) = 40e(0.05)t + C ⇐⇒ P (t) = 40 + Ce−(0.05)t .
The initial condition 1 = P (0) = 40 + C gives that C = −39. Thus,
the amount of Çömezium at any time t > 0 is given by
P (t) = 40 − 39e−(0.05)t .