Register to reply 
A Tricky Problemby Hyperreality
Tags: tricky 
Share this thread: 
#1
Nov1004, 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 Doperator 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. 


#2
Nov1004, 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 


#3
Nov1004, 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



#4
Nov1004, 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"µ",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. 


#5
Nov1004, 01:17 PM

Sci Advisor
HW Helper
P: 11,948

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 "µ" and "m"are different) coefficients,call them A and B.Who stops you from chosing [tex] A=\frac{1}{\mum};B=A [/tex] or viceversa,to find your solution without making any derivatives of the solution given??????? Bonehead... 


Register to reply 
Related Discussions  
My tricky wave problem. ehh its probably not tricky.  Introductory Physics Homework  15  
Tricky Problem...  Introductory Physics Homework  1  
Tricky Problem (at least for me)  Introductory Physics Homework  1  
A very tricky problem that im just not getting  Introductory Physics Homework  2  
Tricky problem  Advanced Physics Homework  2 