LAST (family) NAME: Test # 1 FIRST (given) NAME: Math 2C03

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LAST (family) NAME:
Test # 1
FIRST (given) NAME:
Math 2C03
ID # :
Tutorial # :
Instructor:
Dr. J.-P. Gabardo
Instructions: You must use permanent ink. Tests submitted in pencil will not be considered
later for remarking. This exam consists of 9 problems on 12 pages (make sure you have all
12 pages). The last two pages are for scratch or overflow work. The total number of points is
50. Do not add or remove pages from your test. No books, notes, or “cheat sheets” allowed.
The only calculator permitted is the McMaster Standard Calculator, the Casio fx 991.
GOOD LUCK!
MATH 2C03, SAMPLE TEST 1: SOLUTIONS
#
Mark
1.
2.
3.
4.
5.
6.
7.
8.
9.
TOTAL
Continued. . .
Test # 1 / Math 2C03
-2-
NAME:
ID #:
PART I: Multiple choice. Indicate your choice very clearly. There is only one correct
answer in each multiple-choice problem. Circle the letter (a,b,c,d or e) corresponding to your
choice. Ambiguous answers will be marked as wrong.
1. (4 pts.) Let y(x) be the unique solution to the initial value problem
½ 0
2
y = (x − 2) ey ,
y(0) = 1.
Then, the value of the second derivative of y(x) evaluated at x = 0 is:
(Hint: Do not try to solve the D.E.)
(a) y 00 (0) = 1 + 4 e2
(b) y 00 (0) = −3 e
(c) y 00 (0) = 2 + 6 e
(d) y 00 (0) = 4 e + 3 e2
→ (e) y 00 (0) = e + 8 e2
Solution. We have
2
2
2
2
y 00 = ey + (x − 2) (2 y y 0 ) ey = ey + (x − 2) (2 y (x − 2) ey ) ey
2
2
= ey + (x − 2)2 (2 y e2 y )
and thus, since y(0) = 1,
y 00 (0) = e1 + (−2)2 (2 e2 ) = e + 8 e2 .
Continued. . .
2
-3-
NAME:
2.
(4 pts.)
ID #:
Given that y1 (x) = cos(ex ) is solution of the 2d order homogeneous linear
differential equation
y 00 − y + e2 x y = 0,
use the method of reduction of order to find a 2d linear independent solution y2 (x).
(a) y2 (x) = cos(ex ) ex
→ (b) y2 (x) = sin(ex )
(c) y2 (x) = x cos(ex )
(d) y2 (x) = ex cos(ex )
(e) y2 (x) = ex sin(ex )
Solution.
Using the formula
Z (
y2 (x) = y1 (x)
exp(−
Rx
x0
p(u) du
y12 (x)
)
dx
with p(x) = −1 and y1 (x) = cos(ex ), we have
¡ Rx
¢
Z
Z
exp
−
(−1)
du
ex
x
x
0
y2 (x) = cos(e )
dx
=
cos(e
)
dx.
cos2 (ex )
cos2 (ex )
Making the change of variables, u = ex , du = ex dx, we have
Z
Z
Z
ex
1
dx =
du =
sec2 u du = tan u + C = tan(ex ) + C.
cos2 (ex )
cos2 (u)
Thus, letting C = 0, y2 (x) = cos(ex ) tan(ex ) = sin(ex ).
Continued. . .
-4-
NAME:
ID #:
3. (4 pts.) Let y(x) be the unique solution of the initial value problem:
½ dy
1
= −y 2 e− y ,
dx
y(0) = 1.
Then, the largest interval centered at x = 0 where the solution is defined and continuous is:
(a) (−∞, ∞)
→ (b) (1 − e, e − 1)
(c) (−e, e)
(d) (−2, 2)
(e) (− ln 2, ln 2)
Solution. The DE is separable and can be written as:
−
We have thus
Z
1 y1 dy
e
= 1.
y2
dx
1 1
− 2 e y dy =
y
Z
1 dx = x + C,
or,
1
e y = x + C.
Since y(0) = 1, we have e1 = 0 + C and C = e. We have thus
1
1
= ln(x + e) or y =
.
y
ln(x + e)
Since ln(1) = 0, y tends to infinity as x → 1 − e from the right. The largest interval centered
at x = 0 where the solution is defined is thus (1 − e, e − 1).
Continued. . .
-5-
NAME:
ID #:
4. (4 pts.) The differential equation
£
¤
dy
=0
−3 x3 y 2 − 3 x2 cos(y) + y 2 + [2 x y − sin(y)]
dx
is not exact, but admits an integrating factor µ(x) which is a function of x only. This
integrating factor is:
→ (a) µ(x) = e−x
(b) µ(x) = ex+2 x
3
2
(c) µ(x) = ex
(d) µ(x) =
√
1 + x2
(e) µ(x) = ecos(x)
Solution. We have M (x, y) = −3 x3 y 2 − 3 x2 cos(y) + y 2 and N (x, y) = 2 x y − sin(y). Thus,
My (x, y) = −6 x3 y + 3 x2 sin(y) + 2 y,
and
Nx (x, y) = 2 y.
Therefore
My − Nx
−6 x3 y + 3 x2 sin(y0
(−3 x2 ) (2 x y − sin(y))
=
=
= −3 x2 .
N
2 x y − sin(y)
2 x y − sin(y)
The integrating factor is thus
¶
µZ
¶
µZ
My − Nx
3
2
dx = exp
−3 x dx = e−x .
µ(x) = exp
N
Continued. . .
-6-
NAME:
ID #:
5. (4 pts.) A 2d order homogeneous differential equation of the form
y 00 + p(x) y 0 + q(x) y = 0,
where p(x) and q(x) are continuous on R, has two linearly independent solutions y1 (x) and
2
y2 (x), where y1 (x) = ex . Using the facts that
2
W (y1 , y2 )(x) = e2 x
and y2 (0) = 1,
compute y2 (1).
(a) y2 (1) = 0
(b) y2 (1) = e2
(c) y2 (1) = 1 + e
(d) y2 (1) = e − 2
→ (e) y2 (1) = 2 e
Solution. We have
¯
¯
¯y1 y2 ¯
0
0
x2 0
x2
2 x2
¯
W (y1 , y2 )(x) = ¯¯ 0
.
0 ¯ = y1 y2 − y1 y2 = e y2 − 2 x e y2 = e
y1 y2
It follows that y2 is a solution of the 1st order linear DE
2
y 0 − 2 x y = ex .
The integrating factor is
µZ
u(x) = exp
We have thus
¶
−2 x dx
2
= e−x .
³ 2 ´0
2
e−x y = 1 and e−x y = x + C
2
2
2
Thus, y2 (x) = x ex + C ex . Since y2 (0) = 1, C = 1 and y2 (x) = (x + 1) ex . In particular,
y2 (1) = 2 e.
Continued. . .
-7-
NAME:
ID #:
6. (4 pts.) Let y(x) be the solution of the initial value problem
½ 0
y + x y = g(x),
y(−1) = 1,
where
(
g(x) =
0, x < 0
x, x ≥ 0.
Assuming that y(x) is a continuous solution, compute y(1) .
(a) y(1) = 0
(b) y(1) = 2 + e1
→ (c) y(1) = 2 − e−1/2
(d) y(1) = e3/2 + 2
(e) y(1) = 2 e
Solution. The DE is a 1st order linear DE. The integrating factor is
¶
µZ
2
x dx = ex /2 .
u(x) = exp
³
´0
2
2
2
For x < 0, we have y ex /2 = 0 showing that y ex /2 = C or y = C e−x /2 , for some constant
C. The condition y(−1) = 1 implies that 1 = C e−1/2 or C = e1/2 . Thus y = e1/2 e−x
2 /2
for
x < 0 and, in particular, y(0) = e1/2 . For x > 0, we have
³
´0
2
x2 /2
ye
= x ex /2
2
2
2
and thus y ex /2 = ex /2 + C1 or y = 1 + C1 e−x /2 , for some constant C1 . Since, by continuity,
2
y(0) = e1/2 , we have e1/2 = 1 + C1 and y(x) = 1 + (e1/2 − 1) e−x /2 , for x ≥ 0. In particular,
y(1) = 1 + (e1/2 − 1) e−1/2 = 2 − e−1/2 .
Continued. . .
-8-
NAME:
ID #:
Part II: Provide all details and fully justify your answer in order to receive credit.
7. (9 pts.) Find the general solution of the first order differential equation
5 x y4
dy
= x5 + 5 y 5 ,
dx
x > 0,
using the method of homogeneous equations.
Solution. We can rewrite the DE as
1
dy
=
dx
5
µ ¶4
x
y
+ ,
y
x
which is clearly homogeneous. Letting v = xy , we have y = x v and
of v, the DE becomes
v+x
or,
x
dv
1
= v −4 + v,
dx
5
dv
1
= v −4
dx
5
which is separable. Hence
5 v4
dv
1
= ,
dx
x
and the solution is given by
Z
Z
1
4
5 v dv =
dx or v 5 = ln(x) + C.
x
Hence,
v = (ln(x) + C)1/5 ,
and the general solution in terms of y is
y = x (ln(x) + C)1/5 ,
Continued. . .
C arbitrary.
dy
dx
dv
= v + x dx
. In terms
-9-
NAME:
ID #:
8. (9 pts.) Solve the initial value problem
½ 2 dy
x dx + 2 x y = 2 y 4 x, x > 0,
y(1) = 1
using the method of Bernoulli equations.
Solution. We can rewrite the DE as
x2 y −4
Letting v = y −3 , we have
dv
dx
= −3 y −4
dy
+ 2 x y −3 = 2 x.
dx
dy
dx
and we can rewrite the DE in terms of v as
1
dv
− x2
+ 2 x v = 2 x,
3 dx
which is now a linear DE for v. In standard form, it is written as
dv
6
6
− v=− .
dx x
x
The integrating factor is thus
Z
u(x) = exp(
6
− dx) = exp(−6 ln(x)) = x−6 .
x
Multiplying both sides of the DE by the integrating factor, we obtain
x−6
dv
− 6 x−7 v = −6 x−7
dx
Integrating, we obtain
or
¡
x−6 v
¢0
= −6 x−7 .
Z
x
−6
v=
−6 x−7 dx = x−6 + C.
Since v(1) = (y(1))−3 = (1)−3 = 1, we have 1 = 1 + C, so C = 0. Hence, x−6 v = x−6 and
v = 1. Therefore, y = v −1/3 = (1)−1/3 = 1. Thus, y = 1 is the unique solution of the IVP.
Continued. . .
-10-
NAME:
ID #:
9. (8 pts.) Solve the initial value problem
½ 00
y − 4 y 0 + 13 y = 0,
y(0) = 1, y 0 (0) = 3.
Solution. The auxiliary equation is r2 − 4 r + 13 = 0 and has roots
r =2±
√
4 − 13 = 2 ±
√
−9 = 2 ± 3 i.
The general solution of the DE has thus the form
y = C1 e2 x cos(3 x) + C2 e2 x sin(3 x).
Hence,
£
¤
£
¤
y 0 = C1 2 e2 x cos(3 x) − 3 e2 x sin(3 x) + C2 2 e2 x sin(3 x) + 3 e2 x cos(3 x) .
We have thus y(0) = 1 = C1 and y 0 (0) = 3 = 2 C1 + 3 C2 , from which we deduce that C1 = 1
and C2 = 13 . Therefore, the unique solution of the IVP is
y = e2 x cos(3 x) +
Continued. . .
1 2x
e sin(3 x).
3
NAME:
SCRATCH
-11ID #:
Continued. . .
NAME:
SCRATCH
THE END
-12ID #:

LAST (family) NAME: Test # 1 FIRST (given) NAME: Math 2C03

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