Solving a Tricky Differential Equation Problem with Anton

[Hyperreality]

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.

 

[TenaliRaman]

Ok i haven't 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]

actually you don't have to use L'Hopital, just the plain definition of derivative, which is more elegant i think :P

 

[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.

 

[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.

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 \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??
Bonehead...