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Partial differential equations , rearranging and spotting |
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| Aug27-11, 01:39 AM | #1 |
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Partial differential equations , rearranging and spotting
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. |
| Aug27-11, 04:27 AM | #2 |
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| Aug27-11, 08:47 AM | #3 |
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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].
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| Aug27-11, 09:15 AM | #4 |
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Partial differential equations , rearranging and spottingThey are proportional R=k*r--beta , all k's ( constants cancel ) |
| Aug27-11, 10:53 AM | #5 |
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| Aug27-11, 01:27 PM | #6 |
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| Aug29-11, 09:13 PM | #7 |
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I have been given an equation
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