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à 4.2è Reduction ï Order Given One Solution
äèUse reduction ï order ë fïd ê second fundamental
èèèèèèèsolution ç ê given differential equation.
â For ê differential equation
y»» - 6y» + 9 = 0
one solution isè
eÄ╣
By usïg reduction ï order, ê second, ïdependent solution
is shown ë be
xeÄ╣.
éS The method ç REDUCTION IN ORDER is due ë D'ALEMBERT.
It will be illustrated for a general, lïear, homogeneous,
second order, diiferential equation but can be extended ë
ë differential equations ç higher order for which one or more
solution(s) are known.
Let ê known solution ç
y»» + p(x)y» + q(x)y = 0
be y¬ å construct ê function
y(x) = v(x)y¬(x)
Differentiatïg yields
y»è=èvy¬» + v»y¬
å
y»» =èvy¬»» + v»y¬» + v»y¬» + v»»y¬
èè=èvy¬»» + 2v»»y¬» + v»»y¬
Substitutïg ïë ê differential equation
è vy¬»» + 2v»»y¬» + v»»y¬ + p(vy¬» + v»y¬) + qvy¬ = 0
è vy¬»» + 2v»»y¬» + v»»y¬ + pvy¬» + pv»y¬ + qvy¬ = 0
è v»»y¬ + v»(2y¬» + py¬) + v(y¬»» + py¬» + qy¬) = 0
The coefficient ç v is just ê differential equation that
y¬ was assumed ë a solution so that term is zero å ê
REDUCED EQUATION IN v becomes
v»»y¬ + (2y¬» + py¬)v» = 0
èè This differential equation is a FIRST ORDER one for
ê variableè v»è which is BOTH LINEAR å SEPARABLE å hence
can be solved by ê methods ç Chapter 2.èOnceèv» is known,
it can be ïtegrated directly ë fïd v å hence ë fïd ê
second solutionè
y½ = vy¬
èèThis method is known as reduction ï order because, ê
differential equations that have ë be solved are one order
less than ê origïal.
1 y»» - 5y» + 6y = 0 given eÄ╣ is a solution
A) 1 B) eúÄ╣ C) eì╣ D) eÄ╣
ü With eÄ╣ given as one solution, reduction ï order,
suggests ê oêr solution will be ç ê form
y = veÄ╣
Differentiatïg
y» = v»eÄ╣ + 3veÄ╣
å
y»» = v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣
Substitutïg ïë ê differential equation yields
v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣ - 5(v»eÄ╣ + 3veÄ╣) + 6veÄ╣ = 0
Rearrangïg ï descendïg order ç derivatives ç v
v»»eÄ╣ + v»eÄ╣(6 - 5) + veÄ╣(9 - 15 + 6) = 0
This yields
v»»eÄ╣ + v»eÄ╣ = 0
Cancellïg eÄ╣ yields
v»» + v» = 0
Treatïg v» as ê variable yields
dv»
───è=è-v»
dx
This is a separable first order differential equation
dv»
───è=è-dx
v»
Integratïg both sides yields
ln[v»] = -x
Solvïg for v»
v» = eú╣
Integratïg yields
v = -eú╣
The second solution is ên
y = -eú╣(eÄ╣)è=è-eì╣
As ê sign is taken up ï ê arbitrary constant
y = eì╣
The general solution is
C¬eÄ╣ + C½eì╣
ÇèC
2è y»» - 6y» + 9y = 0ègiven eÄ╣ is a solution
A) eúÄ╣ B) xeÄ╣ C) x D) eÄ╣
ü With eÄ╣ given as one solution, reduction ï order,
suggests ê oêr solution will be ç ê form
y = veÄ╣
Differentiatïg
y» = v»eÄ╣ + 3veÄ╣
å
y»» = v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣
Substitutïg ïë ê differential equation yields
v»»eÄ╣ + 6v»eÄ╣ + 9veÄ╣ - 6(v»eÄ╣ + 3veÄ╣) + 9veÄ╣ = 0
Rearrangïg ï descendïg order ç derivatives ç v
v»»eÄ╣ + v»eÄ╣(6 - 6) + veÄ╣(9 - 18 + 9) = 0
This yields
v»»eÄ╣è=è0
Cancellïg eÄ╣ yields
v»»è=è0
This can be directly ïtegrated twice ë yield
vè=èxè (let ê first constant ç ïtegration = 0)
The second solution is ên
y = xeÄ╣
The general solution is
C¬eÄ╣ + C½xeÄ╣
ÇèB
3 xìy»» - 3xy»è=è0èx > 0ègiven 1 is a solution
A) xî»Å B) xúî»Å C) xÅ D) xúÅ
ü With 1 given as one solution, reduction ï order,
suggests ê oêr solution will be ç ê form
y = v(1) = v
Differentiatïg
y» = v»
å
y»» = v»»
Substitutïg ïë ê differential equation yields
xìv»»- 3xv» = 0
Treatïg v» as ê variable yields
èdv»
xì───è=è3xv»
èdx
This is a separable first order differential equation
dv»èè 3 dx
───è=è────
v» èx
Integratïg both sides yields
ln[v»] = 3 ln[x] = ln[xÄ]
Solvïg for v»
v» = xÄ
Integratïg yields
v = xÅ/4
The second solution is ên
y = xÅ/4(1)è=èxÅ/4
As ê 4 is taken up ï ê arbitrary constant
y = xÅ
The general solution is
C¬ + C½xÅ
ÇèC
4 xìy»» - 2xy» + 2y = 0èx > 0;ègiven x is a solution
A) 1 B) x C) xì D) xÄ
ü With x given as one solution, reduction ï order,
suggests ê oêr solution will be ç ê form
y = v(x) = vx
Differentiatïg
y» = v»x + v
å
y»» = v»»x + 2v»
Substitutïg ïë ê differential equation yields
xì(v»» + 2v») - 2x(v»x + v) + 2vx = 0
Rearrangïg ï descendïg derivatives ç v
xìv»» + (2xì - 2xì)v» + (-2x + 2x)v = 0
Or
xìv»»è=è0
Cancellïg xì gives
v»»è=è0
Integratïg twice (first constant ç ïtegration set ë zero)
vè=è x
The second solution is ên
y = x(x)è=èxì
The general solution is
C¬x + C½xì
ÇèC
5
èèèè xì(x+1)y»» - 2xy» + 2y = 0 given x is a solution
A)èx / [x+1] B) [x+1] / x
C)è x / [xì+1] D) [xì+1] / x
ü With x given as one solution, reduction ï order,
suggests ê oêr solution will be ç ê form
y = v(x) = vx
Differentiatïg
y» = v»x + v
å
y»» = v»»x + 2v»
Substitutïg ïë ê differential equation yields
xì(x+1)(v»» + 2v») - 2x(v»x + v) + 2vx = 0
Rearrangïg ï descendïg derivatives ç v
xì(x+1)v»» + (2xì(x+1) - 2xì)v» + (-2x + 2x)v = 0
Or
xì(x+1)v»» + 2xìv»è=è0
Cancellïg xì gives
(x+1)v»» + 2v» =è0
Treatïg v» as ê variable yields
èè dv»
(x+1)───è=è-2v»
èè dx
This is a separable first order differential equation
dv»èè -2 dx
───è=è─────
v» x + 1
Integratïg both sides yields
ln[v»] = -2 ln[x + 1] = ln[(x+1)úì]
Solvïg for v»
v» = (x+1)úì
Integratïg yields
v = -(x+1)úî
The second solution is ên
y = x(-(x+1)úî)è=è-x / [x+1] or x / [x+1]
The general solution is
èx
C¬x + C½─────
x + 1
ÇèA