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à 1.6èFirst Order Homogeneous (ï x å y) Differential
èè Equations
äèèDetermïe if ê differential equation is
HOMOGENEOUS ï x å y.
â è 2y - xèè
Forèy» =è────── = F(x, y)
è 2x - y
èèèèèè 2(ay) - (ax)èèa(2y - x) èè2y - x
F(ax, ay) =è──────────── =è─────────è=è ──────è=èF(x,y)
èè 2(ax) - (ay)èèa(2x - y) èè2x - y
Therefore, this is homogeneous ï x å y.
éS è A first order differential equation ï ê form
y»è=èF(x, y)
is said ë be HOMOGENEOUS ï x å y if its functional
dependence is on ê ratioèy/xè(or x/y) only å not
on ê value ç x å y ïdividually.
èèTo test for HOMOGENEITY ï x å y, one substitues
ax for x å ay for y ï F.èIf upon simplification, ê
result is ê origïal function i.e. if
F(ax, ay)è=èF(x, y)
ê differential equation is homogeneous ï x å y.
èè è 3x - yèè
Forèy» =è──────è=èF(x, y)
è x + 2y
èèèèèè 3(ax) - (ay)èèa(3x - y) èè3x - y
F(ax, ay) =è──────────── =è─────────è=è ──────è=èF(x,y)
èè (ax) + 2(ay)èèa(x + 2y) èèx + 2y
Therefore, this differential equation is HOMOGENEOUS ï x
å y.
1 èèè3x - y
y» =è──────
èèè x + 2
A) Homogeneous ï x å y
B) Not homogeneous ï x å y
üèè è 3x - yèè
Forèy» =è──────è=èF(x, y)
èèx + 2
èèèèèè 3(ax) - (ay)èèa(3x - y)
F(ax, ay) =è──────────── =è─────────è
èèè (ax) + 2èèèèax + 2
èèAs ê "a" can NOT be facëred å ên canceled,this
differential equation is NOT HOMOGENEOUS ï x å y.
ÇèB
2è èèè3xì - 2xy + yì
y» =è──────────────
èèèè xì - 3yì
A) Homogeneous ï x å y
B) Not homogeneous ï x å y
üèè For èèè3xì - 2xy + yì
y» =è──────────────è= F(x, y)
èxì - yì
èèèèèè 3(ax)ì - 2(ax)(ay) + (ay)ìèè
F(ax, ay) =è──────────────────────────è
èèè è (ax)ì - 3(ay)ì
èèèaì(3xì - 2xy + yì)
è=è ──────────────────
èèèè aì(xì - 3yì)
èèè3xì - 2xy + yì
è=è ──────────────
èèèè xì - 3yì
è=èF(x, y)
Therefore, this differential equation is HOMOGENEOUS
ï x å y.
ÇèA
3 y» =èe╝»╣ tan[x/y]
A) Homogeneous ï x å y
B) Not homogeneous ï x å y
üèè For è F(x,y) =èe╝»╣ tan[x/y]
F(ax, ay) =è e╜╝»╜╣ tan[ax/ay]
è=è e╝»╣ tan[x/y]
è=èF(x, y)
Therefore, this differential equation is HOMOGENEOUS
ï x å y.
ÇèA
4 sï[x + y]
y»è=è ──────────
è x + y
èèèèA) Homogeneous ï x å y
B) Not homogeneous ï x å y
üèè For èèè sï[x + y]
y» =è────────────è= F(x, y)
èx + y
èèèèèèèsï[(ax) + (ay)]èè
F(ax, ay) =è──────────────────è
èèè (ax) + (ay)
èèè sï[a(x + y)]
è=è ───────────────
èèèè a(x + y)
But asèsï[az] ƒèa sï[z], no cancellïg can be done, this
differential equation isèNOT HOMOGENEOUS ï x å y.
ÇèB
äèèFïd ê general solution
â è y» = (2y - x) / (2x - y) is homogeneous.èThe substitution
y = vx, y» = xv» + v yieldsè xv» + v = (2v - 1)/(2 - v)
Rearrangïg yieldsèxv» = (vì - 1)/(2 - v) .èSeparatïg
variable yields ░è2 - v ░èdx
▒ ─────── dvè=è▒è──
▓ vì - 1 ▓è xè Solvïg, via partial
fractions å v = y/x yieldsèy/x + 1 = C(y/x - 1)Äxì
éS èè To solveèèè
y» = F(x, y)
that is HOMOGENEOUS ï x å y, we use ê SUBSITUTION
y = vxèi.e.èv = y/x
Differentiatïg
y» = xv» + v
The homogeneity ï x å y condition converts F(x, y) ë F(v)
so ê substituted equation becomes
xv» + vè=èF(v)
This differential equation is SEPARABLE as
xv»è=èF(v) - v
å
èèdv dx
──────────è=è───
F(v) - v x
è Integratïg both sides will produce an EXPLICIT solution
ï ê variables x å v as long as ê left hå ïtegral
can be done ï terms ç elementary functions.èTo produce ê
GENERAL SOLUTIONèsubsituteèv = y/x.èIn some cases, this can
be solved for an EXPLICIT solution y = g(x).
è In some cases, ê substitutionè x = ywè will produce
easier ïtegration thanèy = vx.
5 3y dx + x dy = 0
A) xyÄ = C B) xìyì = C
C) xÄy = C D) xÅy = C
ü è 3y dx + x dy = 0
can be rearraged ë
è3y
y»è=è- ────è= F(x, y)
è x
èèè 3(ay) èè 3y
F(ax, ay) = - ───────è=è- ────è=èF(x, y)
(ax) èèèx
Thus this differential equation is HOMEGENEOUS ï x å y.
Makïg ê substitutuions
y = vx
y» = xv» + v
yields xv» + v = -3vèè(as v = y/x)
Or xv» = -4v
This separates ë
░è dv ░è dx
▒è────è=è- 4 ▒è────
▓èèv ▓èèx
Integratïg yields
ln[v]è=è-4 ln[x] + ln[C]
or ln[v]è=èln[xúÅ] + ln[C]
Usïg properties ç logarithms yields ê solution ï v
vè=èC/xÅ
Or vxÅ = C
Substitutïg thatè v = y/x yields
(y/x)xÅ + c
so ê general solution is
yxÄ = C
Ç C
6 (xì + yì) dx - 2xy dyè=è0
A) y = xì + Cx B) y = x + Cxì
C) yì = xì + Cx D) yì = x + Cxì
ü è (xì + yì) dx - 2xy dyè=è0
can be rearraged ë
xì + yì
y»è=è ───────è= F(x, y)
è2xy
èè (ax)ì + (ay)ìèèaì(xì + yì)è
F(ax, ay) =è────────────è=è───────────è
èèè 2(ax)(ay)èè aì(2xy)
èèèxì + yì
è=è ───────è=èF(x,y)è
èèèè2xyèèè
Thus this differential equation isèHOMEGENEOUS ï x å y.
Makïg ê substitutuions
y = vx
y» = xv» + v
è 1 + vì
yields xv» + v = ───────èè(as v = y/x)
èè 2v
Or èèè 1 + vìèèèè2v
xv» = ────────è- v ────
èèèè 2v èè 2v
èvì - 1
èè=è- ────────
èè2v
This separates ë
░è2v dv ░è dx
▒è──────è=è- ▒è────
▓èvì - 1 ▓èèx
Integratïg yields
ln[vì - 1]è=è- ln[x] + ln[C]
or ln[vì - 1]è=èln[xúî] + ln[C]
Usïg properties ç logarithms yields ê solution ï v
vì - 1è=èC/x
Or vìx - x =èC
Substitutïg thatè v = y/x yields
(y/x)ìx - x = C
so ê general solution is
yì = xì + Cx
Ç C
7 4x + 3y
y» = - ─────────
2x + y
A) (y + x)(y + 4x) = C
B) (y + x)ì(y + 4x) = C
C) (y + x)(y + 4x)ì = C
D) (y + x)(y + 4x)Ä = C
ü è 4x + 3y
y»è= - ───────è= F(x, y)
`2x + y
èèèè4(ax) + 3(ay)èèè a(4x + 3y)è
F(ax, ay) =è-è─────────────è=è- ──────────è
èèèè2(ax) + (ay)èèèè a(2x + y)
èèè 4x + 3y
è= -è───────è=èF(x,y)è
èèèè2x + yèèè
Thus this differential equation is HOMEGENEOUS ï x å y.
Makïg ê substitutuions
y = vx
y» = xv» + v
èè 4 + 3v
yields xv» + v = - ───────èè(as v = y/x)
èè 2 + v
Or èèèè 4 + 3vèèèè2 + v
xv» = - ────────è- v ───────
èèèèè2 + v èèè 2 + v
èvì + 5v + 4
èè=è- ─────────────
èè 2 + v
This separates ë
░èè 2 + v èèèè░è dx
▒è─────────── dvè=è- ▒è────
▒èvì + 5v + 4 èèèè▓èèx
Usïg partial fraction decomposition ë simplify ê left
ïtegral
èèè2 + v A èèB
────────────────è=è───────è+è───────
(v + 1)(v + 4) èèèv + 1 èv + 4
Cross multiplyïg yields
2 + vè=èA(v + 4) + B(v + 1)
Substitutïg v = -4
-2 = -3Bèi.e.èB = 2/3
Substitutïg v = -1
1 = 3A i.e.èA = 1/3
Substitutïg back ïë ê ïtegral yields
░èè1 èèèè ░è dvèèèèè░è dx
▒è─────èdvè+è▒è─────è=è-3 ▒è────
▓èv + 1 ▓èv + 4èèèè▓èèx
Integratïg yields
ln[v + 1] + 2 ln[v + 4]è=è-3 ln[x] + ln[C]
or ln[v + 1] + ln{[v + 4]ì} =èln[xúÄ] + ln[C]
Usïg properties ç logarithms yields ê solution ï v
xÄ(v + 1)(v + 4)ìè =èC
Substitutïg thatè v = y/x yields
xÄ(y/x + 1)(y/x + 4)ì = C
so ê general solution is
(y + x)(y + 4x)ì = C
Ç C
8 [2x sï[y/x] + 3y cos[y/x]] dx - 3x cos[y/x] dy = 0
A) così[y/x] = CxÄ B) cosÄ[y/x] = Cxì
C) sïì[y/x] = CxÄ D) sïÄ[y/x] = Cxì
ü Rearrangïg
[2x sï[y/x] + 3y cos[y/x]] dx - 3x cos[y/x] dy = 0
yields
è èèè 2x sï[y/x] + 3y cos[y/x]
y»è= ───────────────────────────è= F(x, y)
èè 3x cos[y/x]
èèè2(ax)sï[ay/ax] + 3(ay) cos[ay/ax]
èF(ax, ay) = ──────────────────────────────────è
èè 3(ax)cos[ay/ax]
èèè a{2x sï[y/x] + 3y cos[y/x]}
èè= ──────────────────────────────è
èèa{3x cos[y/x]}
èè=èF(x, y)
Thus this differential equation is HOMEGENEOUS ï x å y.
Makïg ê substitutuions
y = vx
y» = xv» + v
è2sï[v] + 3vcos[v]
yields xv» + v = ──────────────────èè(as v = y/x)
èèè 3cos[v]
Or èèèèè2sï[v]è
xv» + v = ───────è+ v
èèèèè3cos[v]è
èèè 2sï[v]
xv» = ─────────
èèè 3cos[v]
This separates ë
è░è cos[v] èèè ░è dx
3 ▒è──────── dvè=è2 ▒è────
è▓è sï[v] èèè ▓èèx
Usïg substitution on ê left ïtegral
u = sï[v]è du = cos[v] dv
converts that ïtegral ë
è░è duèèè ░è dx
3 ▒è───è=è2 ▒è────
è▓è uèèèè▓èèx
Integratïg
3 ln[u]è=è ln[x] + ln[C]\
Substitutïg back ë v
3 ln{sï[v]} = 2 ln[x] + ln[C]
Usïg properties ç logarithms yields ê solution ï v
sïÄ[v] =èCxì
Substitutïg thatè v = y/x yields ê general solution
sïÄ[y/x] = Cxì
Ç D
äèèSolve ê ïitial value problem
â y» = (2y - x) / (2x - y); y(3) = 2 is homogeneous.èThe substi-
tution y = vx, y» = xv» + v yields xv» + v = (2v - 1)/(2 - v)
Rearrangïg yieldsèxv» = (vì - 1)/(2 - v) .èSeparatïg
variable yields (2 - v) dv / (vì - 1) dx / x.èIntegratïg,
usïg partial fraction decomposition å v = y/x yieldsè
è y/x + 1 = C(y/x - 1)Äxì.è Substitutïg x = 3, y = 2
givesè2/3 + 1 = C(2/3 - 1)Ä3ì i.e. C = -5èy + x = -5(y-x)Ä
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.
9 x dy + 2y dx = 0
y(3) = 2
A) xyì = 18 B) xyì = -18
C) xìy = 18 D) xìy = -18
ü è x dy + 2y dy = 0
can be rearraged ë
è2yèè
y»è=è- ────è= F(x, y)
è x
èèè 2(ay) èè 2y
F(ax, ay) = - ───────è=è- ────è=èF(x, y)
èèèè(ax) èèèx
Thus this differential equation is HOMEGENEOUS ï x å y.
Makïg ê substitutuions
y = vx
y» = xv» + v
yields xv» + v = -2vèè(as v = y/x)
Or xv» = -3v
This separates ë
░è dv ░è dx
▒è────è=è- 3 ▒è────
▓èèv ▓èèx
Integratïg yields
ln[v]è=è-3 ln[x] + ln[C]
or ln[v]è=èln[xúÄ] + ln[C]
Usïg properties ç logarithms yields ê solution ï v
vè=èC/xÄ
Or vxÄ = C
Substitutïg thatè v = y/x yields
(y/x)xÄ + c
so ê general solution is
yxì = C
For ê ïitial condition y(3) = 2, substitute x= 3, y = 2
ë give
2(3)ì = 18 = C
The specific solution is
yxì = 18
Ç C
10 yì dxè-èxì dyè=è0
y(3) = 4
A) è 9y = 7xì + 9x B) è 9y = 7xì - 9x
C) è 9y = -7xì + 9x D) è 9y = -7xì - 9
ü Forè yì dxè-èxì dyè=è0
it rearranges ë
yì
y»è=è────è=èF(x, y)
xì
èèèè(ay)ìèèè aì yìè
F(ax, ay) =èè──────è=è ───────è
èèèè(ax)ìèèè aì xì
èèè yì
è=è ────è=èF(x,y)è
èèè xìèèè
Thus this is a HOMEGENEOUS IN x AND y differential equation.
Makïg ê substitutuions
y = vx
y» = xv» + v
yields xv» + v = vìèè(as v = y/x)
Or xv» = vì - v
This separates ë
░èè 1 è ░è dx
▒è──────èdvè=è ▒è────
▒èvì - v è ▓èèx
Usïg partial fraction decomposition ë simplify ê left
ïtegral
èè1 A èèB
──────────è=è─────è+è───────
v(v - 1) vèèèèv + 1
Cross multiplyïg yields
1è=èA(v + 1) + Bv
Substitutïg v = 0
1 = Aè
Substitutïg v = -1
1 = -B i.e.èB = -1
Substitutïg back ïë ê ïtegral yields
░è 1 èèè ░è dvèèèè░è dx
▒è───èdvè-è▒è─────è=è ▒è────
▓è v èèè ▓èv + 1èèè▓èèx
Integratïg yields
ln[v] -èln[v - 1]è=è ln[x] + ln[C]
Usïg properties ç logarithms yields ê solution ï v
v / (v - 1)è =èCx
Substitutïg thatè v = y/x yields
y/x / (y/x - 1) = Cx
so ê general solution is
(y + x) / x = Cx
or
y + x = Cxì
or
y = Cxì - x
For ê ïitial condition y(3) = 4, substitute x = 3, y = 4
ë give
4 = C(3)ì - 3
or C = -7/9
The specific solution is
9y = -7xì - x
Ç D