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.
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
actually you dont have to use L'Hopital, just the plain definition of derivative, wich is more elegant i think :P
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.
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 "µ" 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... :rofl: