EMCF Quiz #3

MATH 3321
Quiz 3
1. The general solution of y 0 − 2xy = 2xex
2
(a) y = x2 ex + Cex
9/5/14
2
is
2
2
(b) y = 2xex + C
2
(c) y = x2 ex + C
2
(d) y = xex + Cex
2
(e) None of the above.
2. The general solution of x2y 0 − 5xy = −6x3 is
6
5
−3
(b) y = x + Cx−5
(a) y = Cx5 +
(c) y = 2x2 + Cx5
(d) y = −2x3 + Cx6
(e) None of the above.
3. The general solution of
dy
xy 2 − x
=
is
dx
y
(a) ln |y 2 − 1| = ln x + C
(b) ln |y 2 − 1| = Cex
2
(c) y 2 = Cx2 − 1
2
(d) y 2 = Cex + 1
(e) None of the above.
4. The solution of the initial-value problem y 0 + y =
2 ln 2
1 + ex
ln(1 + ex ) ln 2
y=
+ x
ex
e
x e
ln 2
y = ln
− x
x
1+e
e
ex
− ln 4
y=
1 + ex
None of the above.
(a) y =
(b)
(c)
(d)
(e)
1
1
,
1 + ex
y(0) = ln 4
is
5. If y = y(x) is the solution of the initial-value problem xy 0 + 2 y =
ln x
, y(1) = 1,
x
then lim y(x) =
x→∞
(a) 0
(b) 1
(c) 2
(d) Does not exist.
(e) None of the above.
6. The general solution of y ln x y 0 =
(a) y =
√
y2 + 1
is
x
Cx2 − 1
(b) y 2 = C(ln x)2 − 1
(c) y 2 = C(ln x) − 1
(d) y =
q
(ln x)2 + C
(e) None of the above.
7. If y = y(x) is the solution of the initial-value problem xy 0 + y = 2 cos x, y(π) = 0, then
lim y(x) =
x→0
(a) 0
(b) 1/2
(c) 2
(d) ∞
(e) None of the above.
8. The general solution of yex
dy
= e−y + e−2x−y is:
dx
(a) ey = C + e−x + 31 e−3x
(b) yey − ey = C − e−x − 13 e−3x
(c) yey = e−x − 13 e−3x + C
(d) yey + ey = C − e−x + 13 e−3x
(e) None of the above.
2
9. The general solution of (1 + x2 + y 2 + x2y 2 )
dy
= 2xy 2 is
dx
y2 − 1
= ln(1 + x2 ) + C
y
(b) tan−1 y = ln(1 + x2 ) + C
(a)
y2 + 1
= tan−1 x + C
y
y2 − 1
x2 − 1
(d)
=
+C
y
x
(e) None of the above.
(c)
10. The solution of the initial-value problem (y + 1)y 0 = x2y − y,
y(3) = 1 is
(a) ln |y + 1| = 31 x3 − x + 6 − ln 2
(b) y = 13 x3 − x − 5
(c) y + ln |y| = 13 x3 − x − 5
(d) y − ln |y| = 13 x3 − x + 5
(e) None of the above.
11. The general solution of
ex−y
dy
=
is
dx
1 + ex
(a) y = ln [ ln (1 + ex)] + C
(b) y = ln [(1 + ex ) + C]
(c) y = ln [C(1 + ex)]
(d) y = ln [ ln C(1 + ex )]
(e) None of the above.
12. The solution of the initial-value problem xy 0 − y = 2x2 y,
2
(a) y = xex + 1 − e
(b) y = xex
(c) y = ex
2 −1
2 −1
(d) y = xe
x2
−e
+1
(e) None of the above.
3
y(1) = 1 is

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