# A Tricky Problem

by Hyperreality
Tags: tricky
 P: 203 My friend found this problem from Anton Suppose that the auxiliary equation of the equation $$y'' + py' + qy = 0$$ has a distinct roots $$\mu$$ and $$m$$. (a)Show that the function $$g_\mu(x) = \frac{e^{\mu x} - e^{mx} }{\mu - m}$$ is a solution of the differential equation (b)Use L'Hopital's rule to show that $$\lim_{\mu\rightarrow\ m} g_\mu(x) = xe^{mx}$$ 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.
 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
 P: 120 actually you dont have to use L'Hopital, just the plain definition of derivative, wich is more elegant i think :P
HW Helper
P: 11,716

## A Tricky Problem

I's definitely more elegant...
$$\lambda^2+p\lambda+q=0$$
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.
$$\mu^2+p\miu+q=0;m^2+pm+q=0$$
Making the differentiations of "g" correctly and substituting into the original equation,after separating parts with $$\exp{\mux}$$ and $$\exp{mx}$$,will find exactly the 2 equations stated above,which actually will ensure you that g is a solution of th equation.
HW Helper
 Quote by dextercioby I's definitely more elegant... The auxiliary equation reads $$\lambda^2+p\lambda+q=0$$ 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. $$\mu^2+p\miu+q=0;m^2+pm+q=0$$ Making the differentiations of "g" correctly and substituting into the original equation,after separating parts with $$\exp{\mux}$$ and $$\exp{mx}$$,will find exactly the 2 equations stated above,which actually will ensure you that g is a solution of th equation.
2.Why would complicate that much???The general solution to the given ODE is a linear superposition of fundamental solutions,which are $$\exp{\mux}$$ and $$\exp{mx}$$ 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
$$A=\frac{1}{\mu-m};B=-A$$ or viceversa,to find your solution without making any derivatives of the solution given???????