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à 1.5èFirst Order Exact Differential Equations
äèèDetermïe if ê differential equation is EXACT
â è For è xìyÄ dx +èxÄyì dy = 0
(xìyÄ)╤ = 3xìyì
(xÄyì)╨ = 3xìyì
As êse partial derivatives are equal, this is an EXACT first
order differential equation.
éS èèA first order differential equation is said ë be EXACT
if, when written ï ê form
M(x,y) dxè+èN(x,y) dyè=è0,
ê followïg PARTIAL DERIVATIVE equation holds
M╤(x,y)è=èN╨(x,y)
1 3xìy dxè+èxÄ dyè=è0
A) EXACT B) not EXACT
ü Forè3xìy dxè+èxÄ dyè=è0
(3xìy)╤è=è3xì
(xÄ)╨è=è3xì
As êse are equal, ê differential equation is
EXACT.
ÇèA
2 Forèxìy dxè-èxìy dyè=è0
A) EXACT B) not EXACT
ü Forèxìy dxè-èxìy dyè=è0
(xìy)╤è=èxì
(-xìy)╨è=è-2xy
As êse are NOT equal, ê differential equation is
NOT EXACT.
ÇèB
3 cos[x]cos[y] dxè-èsï[x]sï[y] dyè=è0
A) EXACT B) not EXACT
ü Forècos[x]cos[y] dxè-èsï[x]sï[y] dyè=è0
(cos[x]cos[y])╤è=è-cos[x]sï[y]
(-sï[x]sï[y])╨è= -cos[x]sï[y]
As êse are equal, ê differential equation is
EXACT.
ÇèA
4 ye╣╝è+èy(x+1)e╣╝ y»è=è0
A) EXACT B) not EXACT
ü Forèye╣╝è+èy(x+1)e╣╝ y»è=è0
(ye╣╝)╤è=èe╣╝ + y(xe╣╝) = (1 + xy)e╣╝
(y(x + 1)e╣╝)╨è= ye╣╝ + y(x + 1)xe╣╝
= [1 + x + xì]ye╣╝
As êse are NOT equal, ê differential equation is
NOT EXACT.
ÇèB
äèèFïd ê general solution
â è Forè2xy dx + (xì + yì) dy = 0, (2xy)╤ = 2x = (xì + yì)╤
so it is EXACT.èIntegratïg 2xy partially with respect ë x
yieldsèg(x,y) = xìy + h(y).èDifferentiatïg partially with
respect ë y å equatïg ë xì + yìè=èxì + h»(y).
Thusèh»(y) = yì.èIntegratïg this yieldsèh(y) = yÄ/3
The general solution is
xìy + yÄ/3 = C
éSè Letè g(x,y) =èCèè
be ê general solution ç ê EXACT differential equation
M(x,y) dxè+èN(x,y) dyè=è0
Usïg ê chaï rule ë differentiate ê solution yields
è dy
g╨(x,y) + g╤(x,y) ────è=è0
è dx
or splittïg ïë differentials
g╨(x,y) dxè+èg╤(x,y) dyè=è0
As long as g(x,y) satisfies ê usual differentiability
conditions, ê EQUALITY OF CROSS PARTIAL DERIVATIVES requires
that
g╨╤(x,y) = g(x,y)╤╨
è
This can be translated, ï terms ç ê differential
equation ë be solved, ë give ê requirement that ê
differential equation be exact only if
M(x,y)╤ = N(x,y)╨
èèThis also leads ë ê two step process for solvïg an
EXACT differential equation.
èèFirst, asèM(x,y) is ê partial derivative with respect
ë x ç g(x,y) we can partially ïtegrate M(x,y) with respect
ë x ë give
è░
1) g(x,y) =è▒èM(x,y) dxè+èh(y)
è▓
It should be noted that ïstead ç ê usual constant ç
ïtegration from a ïtegration ç a function ç a sïgle
variable, êre is a function ç y alone i.e.èh(y).èThis
is necessary as a partial differentiation with respect ë x
will differentiate any function ç y alone ë zero.
è The second step is ë evaluate h(y).èDifferentiatïg
Equation 1)èpartially with respect y ë yield
èè┤è ░
g(x,y)╤è=è──è▒ M(x,y) dxè+èh'(y)
èè┤yè▓
By ê defïition ç an exact equation, it is also true
that
g(x,y)╤è=èN(x,y)
Equatïg êse two expression å rearrangïg yields a
differential equation forèh(x)
èè ┤è ░
h»(y)è=èN(x,y)è-è──è▒èM(x,y) dx
èè ┤xè▓
This differential equation is generally easily solved by
partial ïtegration with respect ë y å will yield ê
general solution
è è░
g(x,y) =è▒èM(x,y) dxè+èh(y)
è▓
The constant ç ïtegration will be part ç h(y)
èè Consider ê differential equation
2xy dxè+è(xì + yì) dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (2xy)╤ = 2x = (xì + yì)╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒è2xy dxè+èh(y)
è▓
èèè =è xìy + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
xì + h»(y)è=èxì + yì
Thus
h»(y) = yì
So ░
h(y) =è▒èyì dy
▓
èè =èyÄ/3
Thus ê general solution is
xìy + yÄ/3 = C
èèIt should be noted that ê two step procedure could have
been done ï ê opposite order i.e. first partially ïte-
gratïg with respect ë x å ên with respect ë y.èIn some
cases, ê reversal ç order may make for easier computations.
5è -y dx + (yì - x) dyè=è0
A) xy + xÄ/3 = C
B) xy - xÄ/3 = C
C) -xy + xÄ/3 = C
D) -xyì - xÄ/3 = C
üè For -y dx + (yì - x) dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (-y)╤ = -1 = (yì - x)╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒è-y dxè+èh(y)
è▓
èèè =è -xy + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
-x + h»(y)è=èyì - x
Thus
h»(y) = yì
So ░
h(y) =è▒èyì dy
▓
èè =èyÄ/3
Thus ê general solution is
-xy + yÄ/3 = C
ÇèC
6 (xì - yì)è-è2xy y»è=è0
A) xÄ/3 + xyì = C
B) xÄ/3 - xyì = C
C) xÄ/3 + xìy = C
D) xÄ/3 - xìy = C
üè For (xì - yì)è-è2xy y»è=è0
Rearrangïg
(xì - yì) dxè-è2xy dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (xì - yì)╤ = -2y = (-2xy)╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒è(xì - yì) dxè+èh(y)
è▓
èèè =è xÄ/3 - xyì + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
-2xy + h»(y)è=è-2xy
Thus
h»(y) = 0
So ïtegratïg partially with respect ë y will brïg NO
new parts ç ê solution å.
Thus ê general solution is
xÄ/3è- xyì = C
ÇèB
7 (2x + 3y + 4) dxè+è(3x - 2y - 1) dyè=è0
A) xì + 3xy + 4x + yì + y = C
B) xì + 3xy + 4x - yì - y = C
C) -xì + 3xy - 4x + yì + y = C
D) -xì + 3xy - 4x - yì - y = C
üè For (2x + 3y + 4) dxè+è(3x - 3y - 1) dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (2x + 3y + 4)╤ = 3 = (3x - 2y - 1)╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒è(2x + 3y + 4) dxè+èh(y)
è▓
èèè =è xì + 3xy + 4x + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
3x + h»(y)è=è3x - 2y - 1
Thus
h»(y) = -2y - 1
So ░
h(y) = ▒ -2y - 1èdy
▓
èè =è-yì - y
Thus ê general solution is
xì + 3xy + 4x - yì - y = C
ÇèB
8 eú╣sï[y] dxè-èeú╣cos[y] dyè=è0
A) e╣cos[y] = C
B) e╣sï[y] = C
C) eú╣cos[y] = C
D) eú╣sï[y] = C
üè For eú╣sï[y] dxè-èeú╣cos[y] dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (eú╣sï[y])╤ = eú╣cos[y] = (-eú╣cos[y])╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒èeú╣sï[y] dxè+èh(y)
è▓
èèè =è -eú╣sï[y] + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
-eú╣cos[y] + h»(y)è=è-eú╣cos[y]
Thus
h»(y) = 0
So ïtegratïg partially with respect ë y will brïg NO
new parts ç ê solution.
Thus ê general solution is
eú╣sï[y] = C
ÇèD
è9 (4xÄyÄ + 1/x) dxè+è(3xÅyì - 1/y) dyè=è0
A) xÅyÄ + ln[x/y] = C
B) xÅyÄ + ln[y/x] = C
C) xÄyÅ + ln[x/y] = C
D) xÄyÅ + ln[y/x] = C
üè For (4xÄyÄ + 1/x) dxè+è(3xÅyì - 1/y) dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (4xÄyÄ + 1/x)╤ = 12xÄyì = (3xÅyì + 1/y)╨ = N╨
Integratïg M partially with respect ë x
è░
g(x,y) =è▒è(4xÄyÄ + xúî) dxè+èh(y)
è▓
èèè =è xÅyÄ + ln[x] + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
3xÅyì + h»(y)è=è3xÅyì - 1/y
Thus
h»(y) = - 1/y
So ïtegratïg partially with respect ë y
h(y) = - ln[y]
Thus ê general solution is
xÅyÄ + ln[x] - ln[y] = C
Usïg properties ç logarithms
xÅyÄ + ln[x/y]è=èC
ÇèA
äèèSolve ê ïitial value problem
â For ê exact, first order ïitial value problem
è 2xy dx + (xì + yì) dy = 0èè y(2) = 3
This ïtegrates partially ëèxìy + h(y).è
Differentiatïg partially by y requiresèh»(y) = yì.
Integratïg partially yields ê general solution
è xìy + yÄ/3 = C.è Substitutïg x= 2 å y = 3
makesè2ì3 + 3Ä/3 = 21 = C i.e. xìy + yÄ/3 = 21
éS èèA full discussion ç Initial Value Problems for FIRST
ORDER DIFFERENTIAL EQUATIONS is ï Section 1.2.è
èèBriefly, solvïg an Initial Value Problem is a two-step
process.èFirst, fïd ê GENERAL SOLUTION ç ê differential
equation.è Second, substitute ï ê ïitial value ïfor-
mationèi.e.èx╠ for x å y╠ for y.èThis will produce an
equation for C which provides ê value ç ê arbitrary
constant ë put back ï ê general solution.
10 (x - y) dx -èx dyè=è0
y(4) = 7
A) xì + xy = 40
B) xì + xy = -40
C) xì - xy = 40
D) xì - 2xy = -40
üè For (x - y) dx -èx dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (x - y)╤ = -1 = (-x)╨ = N╨
Integratïg M partially with respect ë x
è░
g(x,y) =è▒èx - y dxè+èh(y)
è▓
èèè =è xì/2 - xy + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
-x + h»(y)è=è-x
Thus
h»(y) = 0
So ïtegratïg partially with respect ë y will brïg NO
new parts ç ê solution.
Thus ê general solution is
xì/2 - xy = C
Substitutïgèx = 4 å y = 7èyields
4ì/2 - 4(7) = -20 = C
The specific solution is
xì/2 - xy = -20
or
xì - 2xy = -40
ÇèD
è11 y cos[x] dxè+ (sï[x] - secì[y]) dyè=è0
y(π/2) = π/4
A) y sï[x] + tan[y] =è1 - π/4
B) y sï[x] + tan[y] =èπ/4 - 1
C) y sï[x] - tan[y] =è1 - π/4
D) y sï[x] - tan[y] =èπ/4 - 1
üè For y cos[x] dxè+ (sï[x] - secì[y]) dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (y cos[x])╤ = cos[x] = (sï[x] - secì[y])╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒èy cos[x] dxè+èh(y)
è▓
èèè =è y sï[x] + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
sï[x] + h»(y)è=èsï[x] - secì[y]
Thus
h»(y) = - secì[y]
Integratïg partially with respect ë y yields
h(y) =è- tan[y]
Thus ê general solution is
y sï[x]è-ètan[y]è=èC
Substitutïgèx = π/2 å y = π/4èyields
π/4 sï[π/2] - tan[π/4]è= π/4 - 1 = C
The specific solution is
y sï[x]è-ètan[y]è=èπ/4 - 1
ÇèD
12 (x + y - 3) dxè+è(x - y + 2) dyè=è0
y(x) = 6
A) x║ + 2xy + 6y + y║ + 4y = 4
B) x║ + 2xy + 6y - y║ - 4y = 4
C) x║ - 2xy + 6y - y║ + 4y = 4
D) x║ + 2xy - 6y - y║ + 4y = 4
üè For (x + y - 3) dxè+è(x - y + 2) dyè=è0
Verifyïg that this is an exact differential equation
M╤ = (x + y - 3)╤ = 1 = (x - y + 2)╨ = N╨
Integratïg M paratially with respect ë x
è░
g(x,y) =è▒èx + y - 3 dxè+èh(y)
è▓
èèè =è xì/2 + xy - 3x + h(y)
Differentiatïg partially with respect ë y å settïg equal
ë N(x,y) yields
x + h»(y)è=èx - y + 2
Thus
h»(y) = - y + 2
Integratïg partially with respect ë y yields
h(y) =è- yì/2 + 2y
Thus ê general solution is
xì/2 + xy - 3x - yì/2 + 2yè=èC
Substitutïgèx = 2 å y = 6èyields
2ì/2 + 2(6) - 3(2) - 6ì/2 + 2(6) = 2 = C
The specific solution is
xì/2 + xy - 3x - yì/2 + 2y = 2
or
xì + 2xy - 6x - yì + 4y = 4
ÇèD