Download MAP 2302- Differential Equations

MAP 2302- Differential Equations. Sanchez
Exam 1-Self-Assessment
Answers
1. The sum of $12,000 is invested at the rate of 6% per year compounded
continuously. In how many years will the money triple?
dy
Solution : y(t ) : amount of money at time t ,
 0.06y , y (0)  $12000
dt
dy
 0.06dt  ln y  0.06t  c  y  e 0.06 t  c  y  Ce 0.06 t
y
y (0)  12000  12000  Ce 0  y (t )  12000e 0.06 t
ln 3
 18.31 years
0.06
36000  12000e 0.06 t  3  e 0.06 t  0.06t  ln 3  t 
2. Find a curve in the xy plane that passes through the point
whose slope at any po int ( x, y ) is given by 3 
Solution :
 10 
 1,

21 

6y 1

x x
dy
6
1
dy 6
1
 3 y  
 y  3  ( Linear )
dx
x
x
dx x
x
6
v( x )  e
x6
 x dx
6
 e 6 ln x  e ln x  x 6
dy
6
1
dy

 x6 y  x6  3    x6
 6x 5 y  3x 6  x 5
dx
x
x
dx


 

3x 7 x 6
3x 1
 x 6 y  3x 6  x 5  x 6 y 

cy 
  cx  6
7
6
7 6
10 3 1
10 19
10 25 20  25  5
3x 1
5

  c

cc



y
 
21 7 6
21 42
21 42
42
42
7 6 42x 6
1
3. Solve the differenti al equation y    y  2xy 2
 x
This is a Bernoulli ' s DE ,
 1y  2
u  y1 2  y 1 and du  1y 2 dy
dy
du 1
1
 1y  2   y  1y  2   2xy 2  
 u  2x


dx
x
dx x
 
1

dx
1
1 du
1
1 
 v( x )  e x
 e  ln x  

u  2   u  2
x
x dx x 2
x 
1
1
1
 u  2x  c  u  2x 2  cx  y 1  2x 2  cx   2x 2  cx  y 
x
y
2x 2  cx
 2x 2 y  cxy  1

-1-
4. Solve  x 4  y 4  dx  xy 3dy  0 by finding an integratin g factor


4
4
M  x  y , N   xy 3
N
M
  y 3  Not exact .
 4y 3 ,
X
y
1
1  M  N 



N  y X   xy 3
5
ve
  x dx
 4y 3   y 3    5  f ( x0


x
 e  5 ln x 
1
is an integratin g factor.
x5
3

dx  y dy  0

x4

M 4 y 3 N 4 y 3
M 4 y 3  N 4 y 3
 exact

,

 exact 

,


y
y
x5
x 5 X
x5
x 5 X
y4
f 1 y 4
 c( y )
 f ( x, y )  ln x 
 
M
x x x 5
4x 4
y3
y 3 f
 c' ( y )  c' ( y )  0  c( y )  c


N
x4
x 4 y
y4
 c  0 or 4x 4 ln x  y 4  cx 4  0
Therefore the solution is ln x 
4
4x
 1

 5
x

 4
 x  y 4 dx   1
 5


x


 3
 xy dy  0   1 
x



y4
x5
Note : Solve  x 4  y 4  dx  xy 3dy  0 by the technique of homogeneous


coefficien ts .
Solution :
y
, y  vx and dy  vdx  xdv
x
  x 4  x 4 v 4  dx  x 4 v 3 ( vdx  xdv )  0  (1  v 4 )dx  v 4 dx  xv 3dv  0


dx
v4
 dx  xv 3dv  0  v 3dv 
0
 ln x  c  v 4  4 ln x  C
x
4
4
y
    4 ln x  C  y 4  4x 4 ln x  cx 4  4x 4 ln x  y 4  cx 4  0
x
Let v 
-2-
5. Compute the orthogonal trajectories of the one-parameter family of curves
x 2  3xy  y 2  c
Solution:
2x  3y  3xy '2yy '  (3x  2y )y'  2x  3y  y   
2x  3y
3x  2y
 The slope of the orthogonal trajectory is given by y' 
3x  2y
which has
2x  3y
homogeneous coefficien ts .

Let y  vx and dy  vdx  xdv
 (3x  2 vx )dx  ( 2x  3vx )( vdx  xdv )  0  (3  2 v )dx  ( 2  3v ) vdx  ( 2  3v )xdv  0
 (3  2 v  2 v  3v 2 )dx  ( 2  3v )xdv  0   3  3v 2 dx  ( 2  3v )xdv  0


2  3v
dx
3v  2
3dx

dv 
0
dv 
0
x
x
3v 2  3
v2 1
5

A  2
 3v  2
A
B 
U sin g partial fractions decomposit ion 



1
 ( v  1)( v  1) v  1 v  1 
B 
2

1 
 5
 2

3dx
5
1

 2  dv 
 0  ln( v  1)  ln( v  1)  3 ln x  C
v

1
v

1
x
2
2




 5 ln( v  1)  ln( v  1)  6 ln x  C  ln ( v  1) 5 ( v  1)x 6   C  ( v  1) 5 ( v  1)x 6  e C  k


y
y
 (  1) 5 (  1)x 6  k  y  x 5 ( x  y )  k or ( x  y ) 5 ( x  y )  K where K   k
x
x
6. Transform (x-y+5) dx - (2x-y-3)dy=0 to a homogeneous differential equation and and
then separate the variables. Do not solve.
Solution :
 xy 50
  x  8  0  x  8 and y  13. Let x  u  8 and y  v  13

  2x  y  3  0
(u  8  v  13  5)du  ( 2u  16  v  13  3)dv  0  (u  v )du  ( 2u  v )dv  0
Let v  uw , dv  udw  wdu
u  uw )du  ( 2u  uw (udw  wdu )  0  (1  w )du  ( 2  w )udw  ( 2  w )wdu  0
 (1  w  2w  w 2 )du  ( w  2)udw  0
 ( w 2  3w  1)du  ( w  2)udw  0 
w2
w 2  3w  1
-3-
dw 
du
0
u
7. A certain radioactive substance has a half-life of 38 hours. Find how long it takes for
90% of the radioactive substance to be dissipated.
y(t): amount of radioactive substance at time t
y(0): amount of radioactive substance at time t=0
dy
dy
1
y (38 )  y (0), y   ky 
 ky 
 kdt  ln y  kt  c  y  e kt  c  y  Ce kt
2
dx
y
y (0)  Ce 0  C  y (0)  y(t )  y (0)e kt
y ( 0)
1
1
 ln 2
 y (0)e 38k   e 38k  38k  ln  k 
2
2
2
38
ln 2
t
t


t

38
 y (t )  y (0)e 38  y (t )  y (0)e ln 2
 y (t )  y(0)  2 38
 y (38 ) 
If 90 % is dissipated then y (t )  10 %y(0) or y(t ) 
y ( 0)
10
t
t


y ( 0)
1
t
1
38
38

 y ( 0)  2

2

 log 2
 t  38 log 2 0.1
10
10
38
10
ln 0.1
 t  38 
 t  126 .2332666667 6 hours
ln 2
8. Transform the homogeneous differential equation yy'=-x-2y to a variable separable
differential equation. Separate the variables and do not solve.
dy
 x  2y  0 or ydy  xdx  2ydx  0. Let y  vx and dy  vdx  xdv
dx
 vxvdx  xdv  xdx  2 vxdx  0  v( vdx  xdv)  dx  2 vdx
y


 xvdv  v 2  2 v  1 dx  0 

w 1
w2
dw 
v
v 2  2v  1
dv 
dx
v
dx
0
dv 
 0,. Let w  v  1
2
x
x
v  1
dx
1
1
1
1

 0    w  2 dw  dx  0  ln w   ln x  c  ln( xw)   c
x
x
w
w
w

 lnx( v  1) 
1
x
 c  ln( y  x ) 
c
v 1
yx
9. Solve the OD E e x dy  xy 2dx  0, y(0)  1
Solution :
It is se parable
xex dx 
dy
y2

 0  xex dx 
 c  2 and xex  e x 
dy
 y 2  c  xe
x  ex  1  c  0  1  1  c
y
1
1
1
 2 or  xex  e x  2  y 
x
y
y
xe  e x  2
-4-
10. If v(t)=15sint, L=3 henries, R=6 ohms and I(0)=10 amperes, compute
the value of the current at any time t.
I(t)
V
L
R
dI
dI
dI
 RI  V(t )  3  6I  15 sin t 
 2I  5 sin t (Linear)
dt
dt
dt
dI
v(t )  e  2dt  v  e 2t  e 2t
 2e 2t I  5e 2t sin t
dt

 e 2t I  5e 2t sin t  e 2t I  5e 2t sin tdt  e 2t I  e 2t ( 2 sin t  cos t )  c
L
 

 I  2 sin t  cos t  ce  2t
I(0)  10  10  0  1  c  c  11 and I  2 sin t  cos t  11e  2t
11. Find the differential equation (do not solve) of the family of straight lines with equal
slope and y-intercept.
Solution :
Linear mod el : y  mx  b.
m  b  y'  y  y' x  y'  y  y' ( x  1)  y' 
y
x 1
12. Find the family of orthogonal trajectories of the one-parameter family of curves
y 2  cx3
Solution :
y 2  cx3  c 
y2
x3
 y2
2yy '  3cx 2  2yy '  3
 x3

2

 x 2  2yy '  3y  y'  3y

x
2x

The slope of the orthogonal traje ctorie s is give n by y'  
 3ydy  2xdx 
2x
3y
3y 2
  x 2  k  3y 2  2x 2  K whe re K  2k (a family of e llipses)
2
-5-
13. Test the differential equation 3x( xy  2)dx  ( x 3  2y )dy  0 for exactness.
Show the work .
Solution :
 M
2
M  3x 2 y  6x  y  3x

 It is exact.

 N  x 3  2y
 N  3x 2
 x
14. Use the necessary substitutions to transform the following homogeneous coefficients
differential equation to a variable separable differential equation. Solve the equation
x 2  y 2 dx  xydy  0.
Solution : Le t y  vx and dy  vdx  xdv


 
vdv
dx
1
 1  2 v 2 dx  xvdv  0 

 0  ln(1  2 v 2 )  ln x  c
x
4
1  2v 2
 ln(1  2 v 2 )  4 ln x  C  ln (1  2 v 2 )x 4   c  (1  2 v 2 )x 4  e c  k
2
y
 (1  2  )x 4  k  x 2  2y 2 x 2  k
x
 x 2  v 2 x 2 dx  x 2 v( vdx  xdv)  0  1  v 2 dx  v 2 dx  xvdv  0
15. Use the necessary substitutions to transform (x + 2y -4) dx + (-2x –y +5) dy=0 to a
variable separable differential equation.
 x  2y  4  0
 x  2y  4  0

 3x  6  0  x  2, y  1

  2x  y  5  0   4x  2y  10  0
Le t x  u  2 and y  v  1
(u  2  2 v  2  4)du  ( 2u  4  v  1  5)dy  0
(u  2 v )du  ( 2u  v )dv  0 ( Homoge ne ous coe fficie nts)
Le t v  uw , dv  udw  wdu
 u  2uw du  ( 2u  uw )udw  wdu   0  (1  2w )du  ( 2  w )udw  wdu   0


 (1  2w )du  ( w  2)udw  ( w  2)wdu  0  1  w 2 du  ( w  2)udw  0


 w 2  1 du  ( w  2)udw  0 
w2
w2 1
dw 
du
0
u
( w  1) 3 u 2
1
1 1 
du
3
1
3


.
dw


0

ln(
w

1
)

ln(
w

1
)

ln
u

c

ln
c
 2 w  1 2 w  1
u
2
2
w 1


( w  1) 3 u 2
v y 1
 C  ( w  1) 3 u 2  C( w  1) whe re u  x  2, w  
w 1
u x2
3
3
 y 1

 y 1
  y  x  1
 y  x  3

 1 ( x  2) 2  C
 1  
 ( x  2) 2  C

x

2
x

2
x

2



 

 x2 
 y  x  13  c( y  x  3)
-6-
16. A spherical snowball is melting in such a way that the rate of decrease of its volume is
proportional to its surface area. At 10 A.M. its volume is 500 in 3 and at 11 A.M. its volume
is 250 in 3 . When does the snowball finish melting?
Solution:
dV
4
Step 1.
 kS where V  r 3 and S  4r 2 , V (0)  500 in 3 and V(1)  250 in 3
dt
3
2
2
2
 3V 
4
3V
  4 3V  3  kV 3
Step 2. S  4r 2 and V  r 3  r  3
and S  4 3

3
4
 4 
 4 
2
2
1
1

dV
dV
Step 3.
 kS 
 K V 3  V 3 dV  Kdt  3V 3  Kt  c  V 3  C1t  C 2
dt
dt
1
Step 4. V (0)  500 in  3 500  C 2  V 3  C1t  3 500
Step 5. V (1)  250 in 3  3 250  C1  3 500  C1  3 250  3 500
1
 V 3  3 250  3 500 t  3 500
3




Step 6. V  0  0  3 250  3 500 t  3 500  t 
t
32
3 2 1
3 500
3 250  3 2

3 500  3 250
3 250  3 2  3 250
 4.847322102 or 4 hours 50 min 50.36 seconds
Answer : The snowball will finish melting by 2 : 50 : 50 P.M .
17. A 5-lb roast initially at 50F is put into a 375F oven when t=0. The
temperature T(t) of the roast is 125 when t=75 (min). When will the roast be
medium-rare, a temperature of 150F?
dT
 k (375  T), T(0)  50, T(75)  125, T(?)  150
dt
dT
 kdt   ln375  T   kt  c  375  T  e  kt  c   375  T  Ce  kt  T  375  Ce  kt
375  T
T(0)  50  375  C  C  325  T  375  325e  kt
T(75)  125
 125  375  325e  75k
 e  75k 
ln 1.3
ln1.3
t
t
T(t )  375  325e 75
 150  375  325e 75

250
10
ln 1.3
 75k  ln  k 
325
13
75
ln 1.3
ln 1.3
t

225
 325e 75  225  e 75 

ln 1.3
t 13
ln 1.3
13
13
 e 75  
t  ln  t  75(ln )  ln(1.3)  105 .1185784 min ute s
9
75
9
9
-7-
325
18. Solve the following differential equation by finding an appropriate integrating
factor.


6xydx  4y  9x 2 dy  0
Solution : M  6xy and N  4y  9x 2
 M
 6x

  y
 Not e xact
 N  18x
 x
1  N M 
1

 
18x  6x   2  f ( y )

M  x
y  6xy
y
2
e
 y dy
 e 2 ln y  y 2 is an integratin g factor


y 2  6xydx  y 2 4y  9x 2 dy  0  y 2


 6xy3dx  4y 3  9x 2 y 2 dy  0
 M  6xy3 and N  4y 3  9x 2 y 2
 M
 18xy 2
 y

 Exact
 N  18xy 2
 x
f
M
 6xy3  f ( x, y )  3x 2 y 3  c( y )
x
f
N  4y 3  9x 2 y 2 
 9 x 2 y 2  c ( y )
y
 c ( y )  4y 3  c( y )  y 4  c
 f ( x, y )  3x 2 y 3  y 4  c  0
-8-
19. Solve the following IVP differential equation by finding an appropriate
integrating factor.


xdx  x 2 y  4y dy  0, y )4)  0
 M
0

Solution : M  x and N  x 2 y  4y   y
 Not e xact
 N  2xy
 x
1   N M  1

  2xy  0   2y  f ( x )

M  x
y  x
 e
2ydy
2
 e y is an integratin g factor .
2
2


 e y xdx  e y x 2 y  4y dy  0e y
2
2
2
2
 xey dx   x 2 ye y  4ye y dy  0


2
 M
 2xye y

2
2

 M  xey , N  x 2 y  4ye y   y
 Exact
2
 N  2xye y
 x
2
2
f
1
M
 xey  f ( x, y )  x 2 e y  c( y )
x
2
2
2
2
f
 N  x 2 ye y  4ye y 
 x 2 ye y  c ( y )
y
2
2
 c ( y )  4ye y  c( y )  2e y  c
2
1 2 y2
x e  2e y  c  0
2
2
2
1
y ( 4)  0  4 2 e 0  2e 0  c  0  8  2  c  0  c  10
2
2
2
2
2
1
 x 2 e y  2e y  10  0  x 2 e y  4e y  20  0
2
 f ( x, y ) 
2
 e y ( x 2  4)  20
Anothe r way :




xdx  x 2 y  4y dy  0, y ( 4)  0  xdx  x 2  4 ydy  0 

x
x2  4
dx  ydy  0
y2
1
ln( x 2  4) 
 c  ln( x 2  4)  y 2  C  ln( x 2  4)   y 2  c
2
2
2


2
2
y
 x 2  4  e  y  k  x 2  4  ke y e x 2  4  k
e
y
2
x 2  4  20
-9-
20. Solve the Bernoullis differential equation y'-4y  y -1 , y (0) 
3
2
Le t u  y1 1  y 2 and du  2ydy
dy
3
- 8y 2  2 y -1y , y (0) 
dx
2
du

 8u  2  v( x )  e   8dx  e  8x is an int e grating factor .
dx

du
 e  8x
 8e  8x u  2e  8x  e  8x u  2e  8x
dx
1
1
1
 e  8x u   e  8x  c  u   e  8x e 8x  ce8x  y 2    ce8x
4
4
4
2y

y ( 0) 

3
3
1
1
1
    c  c  1  y 2    e 8x  y 2  e 8x 
2
4
4
4
4
 y  e 8x 
1
4
21. Solve the IVP differential equation y '  2 
y
, y ( 2)  3
x
Solution :
1
dy 1
 x dx
 y  2  v( x)  e
 e ln x  x
dx x
dy
x  y  2 x   xy '  2 x  xy  x 2  c  2(3)  4  c  c  2
dx
2
 xy  x 2  2  y  x 
x
-10-