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A Tricky Problem

by Hyperreality
Tags: tricky
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Hyperreality
#1
Nov10-04, 01:56 AM
P: 203
My friend found this problem from Anton

Suppose that the auxiliary equation of the equation

[tex]y'' + py' + qy = 0[/tex]

has a distinct roots [tex]\mu[/tex] and [tex]m[/tex].

(a)Show that the function

[tex]g_\mu(x) = \frac{e^{\mu x} - e^{mx} }{\mu - m}[/tex]

is a solution of the differential equation

(b)Use L'Hopital's rule to show that

[tex]\lim_{\mu\rightarrow\ m} g_\mu(x) = xe^{mx}[/tex]

I tried to proof this using the D-operator method to find the roots, it doesn't seem to work. There seems to be a simpler way of doing this, but I just can't see it.

Any help is appreciated.
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TenaliRaman
#2
Nov10-04, 04:24 AM
P: 646
Ok i havent given this much thought but try this ,
(Assuming that p and q are constant)
Standard results of polynomials,
p = mu + m
q = mu*m

Differentiate the function g(x) and substitute in the original equation to show that it satisfies the equation.

the answer to b is trivial using L'Hospital,
Differentiate numerator and denominator w.r.t mu,
its easy to see that numerator differentiates to xe^(mu*x)
and the denominator is 1.
the limit evaluates to the required one easily.....

-- AI
ReyChiquito
#3
Nov10-04, 11:34 AM
P: 120
actually you dont have to use L'Hopital, just the plain definition of derivative, wich is more elegant i think :P

dextercioby
#4
Nov10-04, 01:08 PM
Sci Advisor
HW Helper
P: 11,948
A Tricky Problem

I's definitely more elegant...
The auxiliary equation reads
[tex]\lambda^2+p\lambda+q=0 [/tex]
If u chose the solution with "+" to be"",and the one with "-" to be "m",then its solutions verify identically the equation above,i.e.
[tex] \mu^2+p\miu+q=0;m^2+pm+q=0 [/tex]
Making the differentiations of "g" correctly and substituting into the original equation,after separating parts with [tex] \exp{\mux} [/tex] and [tex] \exp{mx} [/tex],will find exactly the 2 equations stated above,which actually will ensure you that g is a solution of th equation.
dextercioby
#5
Nov10-04, 01:17 PM
Sci Advisor
HW Helper
P: 11,948
Quote Quote by dextercioby
I's definitely more elegant...
The auxiliary equation reads
[tex]\lambda^2+p\lambda+q=0 [/tex]
If u chose the solution with "+" to be"",and the one with "-" to be "m",then its solutions verify identically the equation above,i.e.
[tex] \mu^2+p\miu+q=0;m^2+pm+q=0 [/tex]
Making the differentiations of "g" correctly and substituting into the original equation,after separating parts with [tex] \exp{\mux} [/tex] and [tex] \exp{mx} [/tex],will find exactly the 2 equations stated above,which actually will ensure you that g is a solution of th equation.
Hold your horses for a while,junior...
1.It's irrelevant "which is which",as long as they are different.
2.Why would complicate that much???The general solution to the given ODE is a linear superposition of fundamental solutions,which are [tex] \exp{\mux} [/tex] and [tex] \exp{mx} [/tex] with arbitrary (hopefully noninfinite,in this case it applies,sice "" and "m"are different) coefficients,call them A and B.Who stops you from chosing
[tex] A=\frac{1}{\mu-m};B=-A [/tex] or viceversa,to find your solution without making any derivatives of the solution given???????
Bonehead...


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