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- à 1.4è Separable First Order Differential Equations
-
- äèèFïd ê general solution
-
- âèè The differential equationè y» =èxìy
- can be rewritten asè dy / yè=èxì dx
- This is a separable differential equation as ê variables
- are on opposite sides ç ê equation.èIntegratïg each side
- with respect ë its variable yieldsèê general solution
- ln[y] = xÄ/3 + ln Cèèèè╨Ä/3
- Solvïg for y yieldsèèèè y = Ce
-
- éS èèA first order differential equation is said ë be
- SEPARABLE if it can be rearranged ë one ç ê forms
- èèè dy
- M(x)è+èN(y) ────è=è0
- èèè dx
- Or
- M(x) dxè= - N(y) dy
-
- The name separable reflects ê fact ê ê left hå side
- ç ê equation is a function ç x alone while ê right
- hå side is a function ç y only.è
-
- A separable differential equation is solved by
- ïtegratïg both sides with respect ë êir variables.
- ░ è░
- ▒èM(x) dxè=è-è▒èN(y) dy
- ▓ è▓
-
- èèA CONSTANT OF INTEGRATION needs ë be added ë one ç ê
- sides.èFrequently, if one or both ïtegrations result ï a
- NATURAL LOGARITHM, ê constant should be written ï ê form
- ln[C].èThe range ç ln[C] is all real numbers so êre is no
- loss ç generality. This term may lead ë a simpler form ç
- ê general solution via properties ç logarithms.
-
- èèOften, ê general solution cannot be simplied ë yield
- an EXPLICIT general solutionèy = F(x, C)èbut it must be
- left as an IMPLICIT SOLUTION between ê variables x å y.
-
- 1 y»è=èx / y
-
-
- A) y = Cx B) y = xì + C
-
- C) yì = xì + C D) yì + xì = C
-
- ü èèSeparatïgè y» = x / yè
-
- yieldsèèè y dyè=èx dx
-
- Integratïg both sides
- ░ èèè░
- ▒èy dyè =è ▒èx dx
- ▓ èèè▓
- yields, by ê power rule
-
- yì / 2è=èxì / 2è+èK
-
- Multiplyïg both sides by 2 å renamïg ê constant ç
- ïtegration yields ê general solution
-
- yìè=èxìè+èC
-
- ÇèC
-
- è2 y»è=èy / x
-
-
- A) y = Cx B) y = xì + C
-
- C) yì = xì + C D) yì + xì = C
-
- ü èèSeparatïgè y» = y / x
-
- yieldsèèè dy / yè=èdx / x
-
- Integratïg both sides
- ░èdy èèè░è dx
- ▒ ────èè=è ▒è────
- ▓è y èèè▓èèx
- yields, by ê differentiation formula
-
- ln[y]è=èln[x]è+èln[C]
-
- Usïg properties ç logarithms, ê general solution is
-
- yè=èCx
-
- ÇèA
-
- 3 y» = cos[x]così[y]
-
- A) tan[y] = cos[x] + C
-
- B) tan[y] = sï[x] + C
-
- C) cot[y] = cos[x] + C
-
- D) cot[y] = sï[x] + C
-
- ü èèSeparatïgè y» = cos[x]così[y]
-
- yieldsèèèè dyè
- èè───────è=ècos[x] dx
- èècosì[y]
-
- Usïg a trig identity, this can be rewritten as
-
- secì[y] dyè=ècos[x] dx
-
- Integratïg both sides
- ░è è░è
- ▒ secì[y] dyè=è ▒ècos[x] dx
- ▓è èèèèè▓
- èè
- yields, by ê differentiation formulas, ê general solution.
-
- tan[y]è=èsï[x]è+èC
-
- ÇèB
-
- è4 x dxè+èye╣ dyè=è0
-
- A) yì / 2è=èxe╣è+èe╣è+èC
- B) yì / 2è=èxe╣è-èe╣è+èC
- C) yì / 2è=èxeú╣è+èeú╣è+èC
- D) yì / 2è=èxeú╣è-èeú╣è+èC
-
- ü èèSeparatïgè x dxè+èye╣ dyè=è0
- è
- yieldsèèè y dyè=è- x eú╣ dx
-
- Integratïg both sides
- ░ èèè ░
- ▒èy dyè =è- ▒èx eú╣ dx
- ▓ èèè ▓
- The y-ïtegral ïtegrates directly by ê power rule but ê
- x-ïtegral requires ïtegration by parts
- u = xèèèèè du = dx
- dv = eú╣ dx v = -eú╣
- è░
- yì / 2è=è- x (-eú╣)è-è▒ eú╣ dx
- è▓
- This ïtegrates by substitution ë yield ê general solution.
-
- yì / 2è=èxeú╣è+èeú╣è+èC
-
- ÇèC
-
- 5 1 + y
- y»è=è───────
- 1 + x
-
- A) 1 + y =èC(1 + x)
-
- B) (1 + y)ì =è(1 + x)ì + C
-
- C) yì = xì + c
-
- D) yì + 2y = xì + 2x + C
-
- ü èèSeparatïgè
- 1 + y
- y»è=è───────
- 1 + x
- è
- yieldsèè dyè èèè dx
- ───────è=è───────
- 1 + y èè 1 + x
-
- Integratïg both sides
- ░èè dy ░èè dx
- ▒è───────è =è ▒è───────
- ▓è 1 + y ▓è 1 + x
-
- Both can be ïtegrated usïg ê denomïaër ï each case as
- ê substitution variable yieldïg
-
- ln[1 + y]è=èln[1 + x]è+èln[C]
-
- Rearrangïg, usïg properties ç logarithms, yields ê
- general solution
-
- 1 + yè=èC(1 + x)
-
- ÇèA
-
- 6 1 + x
- y»è=è───────
- 1 + y
- A) 1 + y =èC(1 + x)
- B) (1 + y)ì =è(1 + x)ì + C
- C) yì = xì + c
- D) yì + 2y = xì + 2x + C
-
- ü èèSeparatïgè
- 1 + x
- y»è=è───────
- 1 + y
- è
- yields è(1 + y) dyè=è(1 + x) dx
-
- Integratïg both sides
- ░èè èèèèèè░èè
- ▒è(1 + y) dyè =è ▒è(1 + x) dx
- ▓è èè▓
-
- Both can be ïtegrated by ê power rule yieldïg
-
- y + yì/2è=èx + xì/2è+èK
-
- Multiplyïg by 2 å renamïg ê constant prodcues ê
- general solution.
-
- yì + 2yè=èxì + 2x + C
-
- ÇèD
-
- äèèSolve ê ïitial value problem
-
- â For ê separable, first order differential equation
- y' =è-sï[x]/yèèy(0) = 5
- This separates ëèèy dy = -sï[x] dxè
- The general solution isèyì/2 = cos[x] + C
- Substitutïg x = 0 ïë ê general solution yields
- 25/2 = 1 + C so ê ïitial value problem's solution
- isèèyì/2 =ècos[x] + 23/2 or yì = 2cos[x] + 23
-
- 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.
-
- 7 y» = yÄ/xì
- y(3) = 2
-
- A) y = ln[x] + 2
- B) yì = x + xyì/12
- C) yÄ = 3/2 xì - 12
- D) yÅ = 4/3 xÄ - 20
-
- ü èèSeparatïgè y» = yÄ/xì
- è
- yieldsèèè dyèèè dx
- èè────è=è────
- èè yÄèèè xì
-
- Integratïg both sides
- ░è dy èèè░è dx
- ▒è────è =è ▒è────
- ▓è yÄèèèè▓è xì
- Both sides ïtegrate by ê power rule ë yield
-
- -1/2yìè=è-1/x + K
-
- Multiplyïg both sides by -2xyì å renamïg ê constant yields
- ê general solution
-
- yìè=èx + Cxyì
-
- Subsitutïgèy = 2 å x = 3 yields
-
- 4è=è3 + C·3·4
-
- So C = 1/12
-
- The specific solution is
-
- yìè=èxè+ xyì/12
- ÇèB
-
- 8 y» = xì/yÄ
- y(3) = 2
-
- A) y = ln[x] + 2
- B) yì = 2x - 2
- C) yÄ = 3/2 xì - 12
- D) yÅ = 4/3 xÄ - 20
-
- ü èèSeparatïgè y» = xì/yÄ
- è
- yieldsèèèyÄ dyè=èxì dx
-
- Integratïg both sides
- ░è èèè░è
- ▒ yÄ dyè =è ▒èxì dx
- ▓èèèèèè ▓è
- Both sides ïtegrate by ê power rule ë yield
-
- yÅ/4è=èxÄ/3 + K
-
- Multiplyïg both sides by 4 å renamïg ê constant yields
- ê general solution
-
- yÅè=è4/3 xÄ + C
-
- Subsitutïgèy = 2 å x = 3 yields
-
- 16è=è4/3 (27) + C
-
- or 16è=è36 + C
-
- So C = - 20
-
- The specific solution is
-
- yÅè=è4/3 xÄè-è20
- ÇèD
-
- 9 cos[2x] dxè+èsï[3y] dyè=è0
- y(π/2) = π/3
-
- A) cos[3y] = 3/2 sï[2x] - 1
- B) cos[3y] = 3/2 sï[2x] + 1
- C) cos[3y] = -3/2 sï[2x] + 1
- D) cos[3y] = -3/2 sï[2x] - 1
-
- ü èèSeparatïgè cos[2x] dxè+èsï[3y] dyè=è0
- è
- yieldsèèè -sï[3y]dyè=ècos[2x]dx
-
- Integratïg both sides
- ░è èè░è
- ▒ -sï[3y] dyè =è ▒ècos[2x] dx
- ▓è èèèèèè▓è
-
- Both sides ïtegrate by subsitutionèu = 3y on ê left å
- w = 2x on ê right å use trig ïtegration formulas ë yield
-
- cos[3y]/3è=èsï[2x]/2 + K
-
- Multiplyïg both sides by 3 å renamïg ê constant yields
- ê general solution
-
- cos[3y]è=è3/2 sï[2x] + C
-
- Subsitutïgèy = π/3 å x = π/2 yields
-
- -1è=èC
-
- The specific solution is
-
- cos[3y]è=è3/2 sï[2x] - 1
- ÇèA
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