# Partial differential equations , rearranging and spotting

by mohsin031211
Tags: constants, partial differential, rearranging, spotting
 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.
Emeritus
PF Gold
P: 9,789
 Quote by 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.
Simply substitute $R=r^\beta$ into the differential equation.
 Math Emeritus Sci Advisor Thanks PF Gold P: 38,886 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 $e^{\beta ln(r)}= e^{ln r^\beta}= r^\beta$.
P: 100

## Partial differential equations , rearranging and spotting

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

They are proportional
R=k*r--beta , all k's ( constants cancel )
Emeritus