Partial differential equations , rearranging and spottingby mohsin031211 Tags: constants, partial differential, rearranging, spotting 

#1
Aug2711, 01:39 AM

P: 9

I have been given an equation : r^2 ( d^2R/dr^2) + 2r(dR/dr)  lambda*R = 0
It says to assume R~ r^β Then i can't seem to spot how from that information we can produce this equation: β(β − 1) rβ + 2β rβ − λ rβ = 0 Any help would be appreciated, thanks. 



#2
Aug2711, 04:27 AM

Emeritus
Sci Advisor
PF Gold
P: 9,789





#3
Aug2711, 08:47 AM

Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 38,886

That is, by the way, an "EulerLagrange" type equation. Each derivative is multiplied by a power of x equal to the order of the derivative. The substitution t= ln(r) changes it to a "constant coefficients" problem. You should remember that for such an equation we "try" a solution of the form [itex]e^{\beta t}[/itex] (although we then find that there are other solutions). With t= ln r, that becomes [itex]e^{\beta ln(r)}= e^{ln r^\beta}= r^\beta[/itex].




#4
Aug2711, 09:15 AM

P: 100

Partial differential equations , rearranging and spottingThey are proportional R=k*rbeta , all k's ( constants cancel ) 



#6
Aug2711, 01:27 PM

P: 100





#7
Aug2911, 09:13 PM

P: 1

I have been given an equation



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