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Partial differential equations , rearranging and spotting

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mohsin031211
#1
Aug27-11, 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.
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Hootenanny
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Aug27-11, 04:27 AM
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Quote Quote by mohsin031211 View Post
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.
Simply substitute [itex]R=r^\beta[/itex] into the differential equation.
HallsofIvy
#3
Aug27-11, 08:47 AM
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Thanks
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That is, by the way, an "Euler-Lagrange" 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].

stallionx
#4
Aug27-11, 09:15 AM
P: 100
Partial differential equations , rearranging and spotting

Quote Quote by mohsin031211 View Post
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.
Set R= constant * r**beta so dR=constant * whatever

They are proportional
R=k*r--beta , all k's ( constants cancel )
Hootenanny
#5
Aug27-11, 10:53 AM
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Quote Quote by stallionx View Post
Set R= constant * r**beta so dR=constant * whatever

They are proportional
R=k*r--beta , all k's ( constants cancel )
Why the constant?
stallionx
#6
Aug27-11, 01:27 PM
P: 100
Quote Quote by Hootenanny View Post
Why the constant?
Well because I thought tilde is for (constant) linear proportionality.
acnes
#7
Aug29-11, 09:13 PM
P: 1
I have been given an equation


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