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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
-
-
-
-
-