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chapter1.2p
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1996-08-13
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à 1.2èèFirst Order Initial Value Problems ç ê form y» = f(x)
äè Fïd ê general solution
â y» = 6x - 5
èè░
y = ▒ 6x - 5èdxè
èè▓
Integratïg usïg ê power rule yields
yè=è3xì - 5x + C as ê general solution.
éSè A FIRST ORDER DIFFERENTIAL EQUATION ç ê form
y» = f(x)
can be INTEGRATED directly ë yield
èèè░
yè=è▒èf(x) dxè+èC
èè èèè▓
where C is a CONSTANT OF INTEGRATION i.e. an arbitrary
constant.èIt is present as ê derivative ç any constant is
zero, êreby losïg any ïformation about its value.èThus,
ï ê reverse process ç ïtegration we cannot specify which
constant, if any, was present prior ë differentiation.
1 y» = 6xì - 4x + 7
A) 12x - 4 + C B) 2xÄ - 2xì + 7x + C
C) 12x + 3 + C D) 2xÄ - 2xì + 7 + C
ü For y» = 6xì - 4x + 7,
you ïtegrate èè░
y = ▒ 6xì - 4x + 7èdx
èè▓
Usïg ê power rule,
y =è6 xÄ/3 - 4 xì/2 + 7 x + C
The general soution is
y = 2xÄ - 2xì + 7x + C
Ç B
2 dy
──è=è2cos[3x]
dx
A) 6sï[3x] + C B) -6sï[3x] + C
C) 2/3 sï[3x] + C D) -2/3 sï[3x] + C
ü Forè dy
èèè èèè──è=è2cos[3x]
èèèdx
you ïtegrate èè░
y = ▒ 2 cos[3x]èdx
èè▓
Usïg substitution
u = 3x, du= 3 dx i.e. dx = du/3
èè èè░
y = ▒ 2 cos[u]èdu/3
èè▓
y =è2/3 sï[u] + C
Substitutïg back ë ê origïal variable, ê general
soution is
y = 2/3 sï[3x] + C
Ç C
3 y» = xe╣
A) xe╣ + C B) xìe╣/2 + C
B) xe╣ + e╣ + C D) xe╣ - e╣ + C
ü Forè y» = xe╣
you ïtegrate èè░
y = ▒ xe╣èdx
èè▓
Usïg ïtegration by parts
u = x du = dx
dv = e╣èv = e╣
è èè░ è
y = ▒ xe╣èdxè
èè▓
èè ░
è=èxe╣è-è▒èe╣èdx
èè ▓ è
The general solution is
y =èxe╣è-èe╣è+èC
Ç D
4 èèè ╨║
y» = xe
è ╨║ ╨║
A) xìeè + C B) eè/2 + C
è ╨Ä/3 è╨ì
C) x║eèè/6 + C D) xeè + C
ü è Forèèèè ╨║
y» = xe
you ïtegrate èè░è ╨ì
y = ▒ xeè dx
èè▓
Usïg substitution
u = xìèèdu = 2x dxèsoèdx = du/2
è èè░ è
y = ▒ xe╗ du/2xè
èè▓
èè 1è░
è=è─è▒èe╗èdu
èè 2 ▓ è
The solution is terms ç u isè e╗/2 + C
Substitutïg back ë ê origïal variable yields ê
general solution
èèè╨║
y =èeè/2è+ C
Ç B
5 y» = sec[x]tan[x]
A) secì[x]tan[x]/2 + C
B) secì[x]tanì[x]/4 + C
C) sec[x] + C
D) sec[x] + C
ü è Forèy» = sec[x]tan[x]
you ïtegrate èè░
y = ▒ sec[x]tan[x] dx
èè▓
directly via a trig diferentiation formula ë yield ê
general solution
y = sec[x] + C
Ç C
äèèSolve ê ïitial value problem
â è Forèèy» = -2x + 5
y(0) =è4
Integratïg, usïg ê power rules yields ê general
solutionèè y = -xì + 5x + C
Substitutïg x = 0 yieldsè 4 = C
Thus ê solution isèy = -xì + 5x + 4
éS The general solution ë a first order differential
equation contaïs one arbitrary constant ç ïtegration
y = F(x) + C
whereèF is ê ANTIDERIVATIVE ç f ï y» = f(x)
This equation represents a FAMILY OF CURVES which are
identical except for ê value ç C.èExcept for possibly one
poït that ê entire family passes through [For example,
ê family y = cx are lïes passïg through ê origï], ê
members ç ê family are mutually exclusive å pass through
all poïts ï ê plane.èThus, specifiyïg one poït that lies
on ê specific solution will be sufficient ïformation ë
decide on ê value ç C.
Formally, an Initial Value Problem for a FIRST ORDER
DIFFERENTIAL EQUATIONèconsists ç two parts
1) A first order differential equation
2) An INITIAL VALUE, generally ï ê form
y(x╙) = y╙
The process for solvïg ê ïitial value problem is
a two-step one.è
First solve ê differential equation ë produce a
general solution with an arbitrary constant ç ïtegration.
Second, substitute ê ïitial value x╙ ïë ê
general solution å equate ê result ë y╙.èThis is one
equation ï ê unknown C.èSolvïg it produces ê solution
ç ê ïitial value problem.
6 y» = 4x - 7
èèè y(0) = 4
A) 2xì - 7x - 4 B) 2xì - 7x
C) 2xì - 7x + 4 D) 2xì - 7x + 7
ü èè Forèèy» = 4x - 7
Direct itegration produces ê general solution
y = 2xì - 7x + C
Substitutïg x = 0 å y = 4 yields
4 = C
Thus ê solution ç ê ïitial value problem is
y = 2xì - 7x + 4
Ç C
7è y» = xì - 4x
y(3) = 5
A) xÄ/3 - 2xì - 4 B) xÄ/3 - 2xì
C) xÄ/3 - 2xì + 5 D) xÄ/3 - 2xì + 14
ü èè Forèèy» = xì - 4x
Direct itegration produces ê general solution
y = xÄ/3 - 2xì + C
Substitutïg x = 3 å y = 5 yields
5 = 9 - 18 + C
So C = 14
Thus ê solution ç ê ïitial value problem is
y = xÄ/3 - 2xì + 14
Ç D
8 y» = 1/xì
y(1) = 4
A) 1/x + 3 B) -1/x + 5
C) 2/xÄ + 2 D) -2/xÄ + 6
ü èè Forèèy» = 1/xìè= xúì
Direct itegration, usïg ê power rule, produces ê general
solution
y = xúî/-1 + C
è= -1/x + C
Substitutïg x = 1 å y = 4 yields
4 = -1 + C
So C = 5
Thus ê solution ç ê ïitial value problem is
y = -1/x + 5
Ç B
9 y» = 6eÄ╣ + 3
y(1) = 2eÄ + 4
A) 6eÄ╣ - 2 B) 6eÄ╣ + 3x + 1
C) 2eÄ╣ + 4 D) 2eÄ╣ + 3x + 1
ü èè Forèèy» = 6eÄ╣ + 3
Splittïg this ïë two ïtegrals å usïg substitution
on ê first yields
èè ░
y =è▒è6eÄ╣ + 3èdx
èè ▓
u = 3xèdu = 3 dxèsoèdx = du/3
░ èè░
yè=è2 ▒ e╗ duè+è▒è3 dx
▓ èè▓
è =è2 e╗è+è3xè+èC
Substitutïg back ë ê origïal variable produces ê
general solution
y = 2 eÄ╣è+è3xè+èC
Substitutïg x = 1 å y = 2eÄ + 4 yields
2eÄ + 4 = 2eÄ + 3 + C
So C = 1
Thus ê solution ç ê ïitial value problem is
y = 2eÄ╣ + 3x + 1
Ç D