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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