Download 1. An integrating factor for the following differential equation 2xy dx +

Mathematics 2C03 Test 1
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MULTIPLE CHOICE
There are 3 multiple choice questions. All questions in this section have the same value. A
correct answer scores 4 and an incorrect answer scores zero. Record your answer by circling ONE
and ONLY ONE of the letters. Ambiguous answers will be considered as incorrect.
1. An integrating factor for the following differential equation
2xy dx + (y 2 − 3x2 ) dy = 0
is:
(A) ey y 2
(B) ey y −2
(C) y −4
(D) y 2
(E) 2y 2 + y 4
SOLUTION: We look for an integrating factor of the form µ(y) so that
∂
(2xy u(y)) = 2xy µ0 (y) + 2x µ(y)
∂y
is equal to
∂
(µ(y) (y 2 − 3x2 )) = −6xµ(y).
∂x
0
But then, 2xy µ (y) + 2x µ(y) = −6x µ(y) implies that
µ0 (y) = −4
y µ(y), a separable DE. Solving we find an integrating factor,
−4
µ(y) = y .
Continued on next page.
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Mathematics 2C03 Test 1
Name:
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2. Let y(x) be the unique solution to the initial value problem
y 0 = y 2 cos(y),
y(0) = a.
For which one of the values of a listed below, does the solution y(x) satisfy
lim y(x) = 0?
x→+∞
(Hint: Consider the phase line. Do not try to solve the ODE.)
(A) a = − π2
(B) a = − π4
(C) a =
(D) a =
(E) a =
π
4
π
2
3π
4
SOLUTION: Any value where − π2 < a < 0. Therefore, a = − π4 .
3. Evaluate the Wronskian W (y1 , y2 , y3 )(0) where
y1 (x) = e−t , y2 (x) = et , y3 (x) = e2t .
(A) 0
(B) 2
(C) 4
(D) 6
(E) 8
SOLUTION: Let r1 = −1, r2 = 1, and r3 = 2.
W (y1 , y2 , y3 )(t) = er1 t er2 t er3 t Π1≤i<j<3 (rj − ri ).
Therefore,
W (y1 , y2 , y3 )(0) = (r3 − r1 )(r3 − r2 )(r2 − r1 ) = (2 − (−1))(2 − 1)(1 − (−1)) = 6.
Continued on next page.
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Mathematics 2C03 Test 1
Name:
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COMPLETE ANSWER QUESTIONS
Questions 4-6 You must show your work to to receive full credit. You will be
graded on the clarity and presentation of the solution, not just upon whether or not
you obtain the correct solution. The value of each question is given in the margin.
[14 ]
4. In this question marks will only be given for the justification.
Without solving, state with justification whether the differential equation
2u3 dv + v(6u2 − 2v 2 ) du = 0
is:
[2 ]
(a) separable.
f and g.
NO, since
2
(b) homogeneous.
[2 ]
(c) exact.
[2 ]
(d) linear in u.
NO, since
du
dv
[2 ]
(e) linear in v.
NO, since
dv
du
[2 ]
(f ) Bernoulli in u.
NO, see (d).
[2 ]
(g) Bernoulli in v.
YES, see (e).
Continued on next page.
YES, since
dv
du
[2 ]
NO, since
2
= − v(3uu3−v ) =
6 f (u)g(v) for some functions
du
dv
∂
3
∂u 2u
= −(3 uv − ( uv )3 )
= 6u2 and
=
∂
2
∂v v(6u
− 2v 2 ) = 6u2 − 6v 2 .
−u3
v(3u2 −v 2 ) .
+ u3 v =
1 3
u3 v .
Page 4 of 6
Mathematics 2C03 Test 1
Name:
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5. Show that the following differential equation is exact and solve the
initial value problem,
[13 marks]
(6xy + 4) dx + (3x2 + 10y) dy = 0,
SOLUTION: It is exact since
∂
∂y (6xy
y(1) = 2.
+ 4) = 6x is equal to
∂
2
∂x (3x
+ 10y) = 6x.
∂
We look for a function F (x, y) where ∂x
F (x, y) = 6xy + 4.
Integrating both sides with respect to x we obtain,
F (x, y) = 3x2 y + 4x + g(y) where g(y) is an arbitrary function of y.
But then,
3x2 + 10y =
∂
∂
F (x, y) =
(3x2 y + 4x + g(y)) = 3x2 + g 0 (y).
∂y
∂y
Therefore, g 0 (y) = 10y and so g(y) = 5y 2 .
Therefore F (x, y) = 3x2 y + 4x + 5y 2 ,
and the general solution is
F (x, y) = C, C an arbitrary constant, i.e.,
3x2 y + 4x + 5y 2 = C.
To find the solution of the initial value problem, since y(1) = 2, it follows that
C = (3)(12 )(2) + 4(1) + 5(22 ) = 30. Therefore, the solution of the initial value
problem is
3x2 y + 4x + 5y 2 = 30.
Continued on next page.
Page 5 of 6
Mathematics 2C03 Test 1
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6. Find the general solution of the first order homogeneous differential
equation
p
y 2 + x x2 + y 2
dy
=
, x > 0.
dx
xy
[11 marks]
SOLUTION: Dividing numerator and denominator by x2 it follows that
p
(y/x)2 + 1 + (y/x)2
dy
=
.
dx
y/x
Consider the change of variables. v = y/x. Then y = xv and
Then we can rewrite the DE,
√
dv
v2 + 1 + v2
x+v =
.
dx
v
dy
dx
=
dv
dx x
+ v.
Subtracting v from both sides we obtain,
√
dv
1 + v2
x=
,
dx
v
a separable DE. Therefore,
√
v
1
dv = dx.
x
1 + v2
Integrating, we obtain
1
(1 + v 2 ) 2 = ln |x| + C.
Replacing v = y/x, it follows that the solution in implicit form is
1
(1 + (y/x)2 ) 2 = ln(x) + C,
Continued on next page.
C an arbitrary constant.
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