Answer Key
15.7 Notes Differential Equations
-
University:
Fresno Pacific University -
Course:
MATH 210 | Calculus I Academic year:
2019
-
Views:
363
Pages:
6
Author:
Nicole Wang
Honors Brief Calc.
Name
Period
15.7 Notes
Differential Equations
In studies of physical, chemical, biological, and other natural phenomena, scientists attempt, on
the basis of long observation, to deduce mathematical laws that will describe and predict nature’s
behavior. Such laws often involve the derivatives of some unknown function F, and it is required
to find this unknown function F.
If given function f, it may be required to find all functions y
= F (x) so that
^
=
OF
X?
f (x)
This equation is an example of what is called a differential equation . A function y = F ( x ) for
solution of
which
= fix)
= f { x ) is a solution of the differential equation. Thei Lgeneral
<
(
i
\
i
' 1
consists of all antiderivatives of f.
^
^
EX 1: Find the general solution of the differential equation
— = 5x
dy
ti A
+ 2~ X 1-
*
= fhf i
'
c
-I
5 - M-5 + fe -i c
c - - */ <,
^.
= 5x 2 + 2 with a boundary
*1
>> ~"
y
Li L>
4
^
-
EX 3: Solve the differential equation below with the boundary condition that y = - 1 when x = 3.
4
«
3
& -v 2.** -4 X
3
^
3^ + 3 + ^
£
3!
3
C
4
4
1
r~
j
^ X3 -
f
y7
j
Y
~ J 9'
,
i i>
-fu ^ c
C
EX 2: Find the particular solution of the differential equation
condition that y = 5 and x = 3.
)
^ Applications in Business, Biology, and Natural Sciences
The amount A of a substance varies with time t in such a way that the time rate of change ofA is
proportional to A itself
Mathematical formulation of this statement:
if k > 0, time rate of change is positive
if k < 0, time rate of change is negative
ft
dA
dt
— = kA
v ayVvovUVr
VVSW vV
M r<\e-
(6
4-
_
'
which is called exponential law , law of
Solution of the differential equation is J Pt
uninhibited growth or decay , or the law of continuously compounded interest
fiMrj
u,
EX 4: Assume that the population of a colony of bacteria increases at a rate proportional to the
number present. If the number of bacteria doubles in 5 hours, how long will it take for the
bacteria to triple?
b&CftoSL pirxnJ cti •
to d A
c
“fet >
\ nvW \
\r\ Z K
0 1 AV /h
-? 9b
h hH
d
a
/A
fto
fir k
p/ r i ( n -f
'
1
"
0
bcuUhui 0 / /
= - Ifr 2.
t = . 13^
•
A < ) - ' /4
f )oC
X
=
time required for half of the substance to decay
fa
%
r
)
Xh % y { * In dy
determined
that about 20% / / EX 5: In the skull of an animal found in an archaeological dig, it was
of the original amount of carbon-14 was still present . If the half-lif of carbon-14 is 5600 years,
6ov\d.f *\ * -
*
=r jCfr A
dt
Half-life -
1K
hive t
-
ft -
^
present
n a \ amt A cm
d / r z )LA
dt
Sbludm is A
"
^
A
0 d
iff
3L *
In -i
)
*
C
Uf 'f
nt
#,00
if Is OQ K~
s f j, 00
P
- t> oa \?M
f)=
fly d 'v'd
UJ %
;
h n>
lc A
Ac e TPTT
p
*
9i h j $ fUvd
Ciumy\ - N
'
iKA
isofl -
Att ", 0°0l i l t
% fa c - 0 0 b I i
4
-trv-e
ft ( -t )*
'
•
Xo-
\ v\
v
-
?
,
e-^ ‘ '
oc >
c
?
,
13^
.
—
jj
jfr
lo%
\+•
[f Slope Fields - Classwork
In this section , we can solve differential equations by obtaining a slope field (sometimes called a direction field) that
approximates the general solution . The slope field of a first- order differential equation says that the differential
equation can be interpreted as a statement about the slopes of its solution curves. You are given dy / dx .
dy
dy
dy 1
dy
Fill in the chart for
i)
= V Fill in the chart for
l
dx
dx
dx
%
j
( x, y )
3
2
1
0
1
-2
-3
-
-3
-2
-1
3
2
1
0
-1
-2
-3
3
2
1
0
-1
-2
-3
3
2
1
0
-1
-2
-3
2
3
2
1
0
-1
-2
-3
1
3
2
1
0
-1
-2
-3
0
3
2
1
0
-1
-2
-3
3
3
2
1
0
-1
-2
-3
%
:pTTpFH TTs^TTfi w P « n 1C
Turn
ZOOM
Moth Dra
{
'/
|
TracelReSrarh
/ / / /
/ / /7 7 7 / / / /
/ / / / / / / /
/
*
S S S S S S S
\
\ \ \ \ \ \ \ \ \ \ \ \ \
v.
V
V
X
X
( i)|4l
DE
Find the solution going through
b) (- 2, 0)
a) (1, 1)
dy
dx
3)
OC-3
_ -x
y
-3
(x , y )
3
1.5
.3
2
1
0
-1
-2
-3
-1
oo
-3
. 67
1
2
. 33
.5
1
0
0
0
0
oo
oo
-2
-1
-.5
-.33
-.1.5 -1
-1
- . 67
-
-
2
-. 33
-. 5
-.67
oo
-1
oo
-1
-2
oo
0
0
0
1
.5
. 33
2
1
. 67
^
"x V
/ /
\ \
\ \
V \
MAIN
S
^
-
RAD E X A C T
—
-1
-1
oo
oo
oo
oo
oo
oo
oo
-.5
-1
-.5
-.5
-. 5
-.5
-1
-1
-1
^\
/
/
^\
V
1
1
1
1
1
2
.5
.5
.5
.5
.5
.5
.5
3
. 333
333
333
333
333
333
333
2
5
4
3
2
1
0
-1
3
6
5
4
3
2
1
0
-
x
-
^
3
-1
- 1.5
-3
oo
3
1.5
1
dy
= x+ y
dx
Fill in the chart for
-3
0
-1
-2
-3
-4
( x, y)
3
2
1
0
-1
-2
-3
-5
-6
- TFFHITI
.
-
V
\
—
^
/ / /
\
/
/ / / /
\ \ \ \ W^
\ \ \ \ \ \ \
V
\ \ \ \
MAIN
\
//
^ -\
N
\ \ \ \ \ \
DE
RAD EXACT
/
/ I I I
/ / / f
- ^^
.
s
/
—
dy
—
dx
1
4
3
2
1
0
-1
-2
0
3
2
1
0
-1
-2
-3
2
1
0
-1
-2
-3
-4
I' FTTYTzTYnj F H Y F S T T FS
Tools ZOOM Trace ReSrap h Math Draw Per IC
S / / / 7 / 777 / 7
r
'
\ \ \ \
Find the solution going through
b) (-2, 0)
a) (1, 1)
-1
-2
1
0
-1
-2
-3
-4
-5
N
DE
DE
RAD EXA
MAIN
4)
1
FH
F 5TT Ffi * TZ’ pQ h
Fi r Y FZ r T F 3
Tools Zoom TKOCO ftO KOPh Moth DKOW Rtn 1 C
f
-1
-.5
1
1
1
Find the solution going through
b) (-2, -1)
0 , 1)
dy
Fill in the chart for
dx
-2
1 St
3
0
\ T/
\
RAD EXACT
MAIN
-1
——
— — ---
V V V >x V > X V
\
\ \
-2
-.5
7
S S S S
- - \ - -\ \
i
. - 33
. - 33
.- 33
. - 33
. - 33
. - 33
.- 33
J
\
t
-3
F H F S f 7s T
Tools Zoom Tro « RoSroph Moth Draw Pen IC
Tools
V
(x , y )
3
2
1
0
-1
-2
-3
- J-
/ /
*
Find the solution going through
b) (- 2, -1)
«) (1, 1)
%
'K
Cy^ fh
In
l Yi 'X Match the slope fields with their differential equations.
(B)
(A)
W T\
\\
\
k
TTTm
/ / / / / / /
/ / //7 / '
t ' f ' f r n\’-fiVi-77
i i/i I I i / / / / / /
LILLJLJLJJLLJJL - I -1 I \
(C)
flX
^V /
-
/
-CM
—dyetc =
j/yi ^
— —— —
sinx
£
\
8.
i i n n
\ \ \ \
\ \ \ \
IQ
x-y
9-
tb
\ \ \
—
(B)
/ /
{ / / / / I I
i i \
m
\
\
—
/ /
t
a
1 0.
,
*
dx = B
-/ / /
bo
/
w/ /
/ / /
-/
o
-^w
*
\
\ \ \ \ \
\ \ \ \ \ \ --/ / y / y y
\ \ \ \ w ^/ / / y y
\ \ \ \ ^
/ /
^
--
\ \ \ \ \ \
d
13.
dy
^
dx
x
y
D
JC~A
^
-^ % \ \ \ \
/ / / / s'
/ I / // '
\ \ \
^
-
7
<7
f
/
\ w
\ \w
\ \ \ \ \ \
dy
12. -^ = 0.5
-
M O
.
\ \ \ \ \
\ \ \ \ \
dx
^
f ( T O c s w*
rrrTTrn=7777
(D)
dy
= 0.5JC -1
c£c
10 o ) ^
t i t i t \
/ *
VWO
I I I I /
11.
/ i n n
/
,
/ /
- s' // // //
777777 777777
/ / / / / / ' / / / / / /
uuu
——
\ \ \ \ \ \W
\ \ V\ \
\l
(C)
—
^
\’ o W 0
\ \ \ \- / / / / / / / i
\ \ \ \ \-/ / n n i
\ \ \ \ \ w-/ / y / y
i \ \ \ \ \ \\ // y y
.
f = 2- y A
Match the slope fields with their differential equations.
(A)
\ \ \
vv v v \vv '
w
dy
f
I I \ \ \ \ \\ // y
I \ \ \ \ \ ,\ // / y
\ \ .\ . \ ..\ V - / / y y y
%
—dx =
f
/
(D)
-
- SV" w> -v /
^
7.
\ \ V I \\ r / / f i
\ \ \ A W / / /I I I I
\ \ \ \\\- - / / I I I I
\\ \ \
\ N - - / /7 / / /
\
7 ‘ I' M
\ \ \ \\\-- / / t i l l
\ \ \ \ \
I i t t t
\ V+ / /
\ 'v
w / /^w^
/" W - / / \ ^
“v '
--
-- -
Is O
w\wW W W
14.
dy
dx
=
jc
_
+y
Ax
= 7C From the May 2008 AP Calculus Course Description:
15.
y
3
/
/ / /
/ / /
I / / /
I / / /
f
/
/
/
s s s
I I
/ / / S S
I
/ / S S S S
I
I I / /
I f / /
h i / /
I I I /
i f
/
I I
I
I
\
I
/
S
/
/
-
/
S
S
s s
s/s
S’
S
/
s/ i s s s
/
/
/
// / / S s
s s
s
s s
S
/
/
/
^
^ ^
s s
/// / / S S s
//
//
/
/ / /
/
/
/ /
/ /
I / /7
I / // /
/
i
/
/ / / s
S
^ ^
/ / / s s s s s ^
s ss
s "
s'
/ / / S’ s’'
y
s
y
s' s' s’ ’ s s'
/ /
- --
0
s
s
3
*~ x
The slope field from a certain differential equation is shown above. Which of the following could be
a specific solution to that differential equation?
(A) y - x2
(B) y =
ex
(C) y - e~ x
(D) y = cos x
(E) y =\nx
XV
16.
T
The slope field for a certain differential equation is shown above. Which of the following could be a
specific solution to that differential equation?
(A) y = sin x
(B) y = cos x
(C) y = x
(E) y = In x Differential Equations by Separation of Variables - Classwork
dy
= / (*) • g ( y ) . In order to solve it, you must put it in the form of
dx
g ( y ) dy = / (*) • dx allowing you to integrate. Your goal is to get an equation in the form of y = h ( x )
dy x + sin ( x )
2
dy 2 x
=u
3y 2
dx
dx
dx y
3 y 2 dy = [x + sin ( x )] dx
y 2dy f d x
f
f
dy
y
dx
f - f 2x
3)
2) y -l
1)
x2
- cosx + C
y
y
x+C
=
2
=x +C
1
2
-1
y
- cosx + C
y = ± V2 x 2 + C
x+C
2
A differential equation will be in the form of
*
_
^
-
'
‘
-f
-
—dxdy = 4
4)
J
dy
= ky
dx
y
f f - f^
|_ | = x
ln y
f
J = fkdx
y = e 4 r +C
y=e
.
6)
fct +C
y = CeLx
4x
J = fxdx
InM - y + C
f
5 . ln|>’| = lex + C
4 +C
y = Ce
dy
= xy
dx
y=e
—dudt = e
7.
u + 21
*
= Ce
2
2
dx
= l + / - x - rx = ( l + r )( l - x )
dt
u
2t
= e •e
-
/ * / *” *
-e
ii2 +c
8.
1
2
2
2
= — e ' + C => it = - ln -- e ' + C
2
i d1 -x / 0
X
“
+
0*
A
-ln|l - x\ = t + — + C
,
2
l-x = e —
,
2
-2
< +C
=> x = 1 + Ce
2
2
/
Find the solution of the differential equation that satisfies the given condition .
xe-t
9.
J
dy
dx
dx
= 1 , x(v0) = 1
dt
xdx =
J
’
edt ==>
_
-
1+ x
, (1) = -4
xy >
2
2
— = er + C
-1
=> x
—21 = 1 + C => C = —
2
2
= 2e ' - 1
t/x =>
10.
—y2 = lnx + x + C
8 = l + C => C = 7 => y 2 = 21nx + 2 x + 14
x + 2 yv/ x 2 + 1
11 .
Jf —
y2 + l
t/y =
J dx => tan ’ y = x + C
"
y = tan ( x + C ) => c + - l => y = tan ( x - 1)
12. 2 Jf y dy =
f . 2* — dx =t
J
>
Vx + 1
l = C - l => C = l => y = 2 2
4
y2 = C - x2 + 1
Vx 2 + l
15.7 Notes Differential Equations
of 6
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