Partial differential equations , rearranging and spotting

[mohsin031211]

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.

[Hootenanny]

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 R=r^\beta into the differential equation.

[HallsofIvy]

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 e^{\beta t} (although we then find that there are other solutions). With t= ln r, that becomes e^{\beta ln(r)}= e^{ln r^\beta}= r^\beta.

[stallionx]

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]

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]

Why the constant?

Well because I thought tilde is for (constant) linear proportionality.

[acnes]

I have been given an equation