Differential eqtn (What's wrong with my calculation?)

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athrun200
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I would like to ask why I obtain a different answer from the book?
If I am wrong, how to obtain the correct answer?

Homework Statement



Printed textbook page showing the derivation of the small-rho limit of a radial differential equation, with equations 4.56 to 4.59 and a footnote; the general solution u(rho) = C rho^(l+1) + D rho^(-l) and the conclusion u(rho) ~ C rho^(l+1) are circled in red.


Homework Equations





The Attempt at a Solution



Handwritten derivation: d2u/dp2 = l(l+1)/p2 u, then characteristic equation r2 = l(l+1)/p2, r = ±√(l(l+1))/p, giving u(p) = Ae^(√(l(l+1))/p · p) + Be^(-√(l(l+1))/p · p), simplified to Ae^√(l(l+1)) + Be^(-√(l(l+1))).
 
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  • Printed textbook page with a highlighted passage solving a differential equation: for small rho, d2u/drho2 = l(l+1)/rho2 u, giving the general solution u(rho) = C rho^(l+1) + D rho^(-l), with D = 0 so u ~ C rho^(l+1).
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  • Handwritten working on lined paper: d²u/dp² = l(l+1)/p² u, then the characteristic equation r² = l(l+1)/p², r = ± √(l(l+1))/p, and u(p) = A e^(√(l(l+1))/p · p) + B e^(−√(l(l+1))/p · p), simplified to A e^(√(l(l+1))) + B e^(−√(l(l+1))).
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Rewrite the differential equation slightly to get
[tex]\rho^2u''(\rho)-l(l+1)u(\rho) = 0[/tex]
Then substitute a trial solution of the form [itex]\rho^r[/itex] and then solve for r.
 
vela said:
Rewrite the differential equation slightly to get
[tex]\rho^2u''(\rho)-l(l+1)u(\rho) = 0[/tex]
Then substitute a trial solution of the form [itex]\rho^r[/itex] and then solve for r.

Is it another way to solve ODE?
What is the name of this method?

I can only remember to sub e^r:-p
 
You basic error is trying to apply a method that works for equations with constant coeficients to an equation with variable coefficients. As a result your characteristic equation, which should not involve the idependent variable, is wrong and everything after that is wrong.
 
HallsofIvy said:
You basic error is trying to apply a method that works for equations with constant coeficients to an equation with variable coefficients. As a result your characteristic equation, which should not involve the idependent variable, is wrong and everything after that is wrong.

Handwritten differential equation: y double prime equals x squared times y.


how about this one?
How to solve it?
 
Attachments
  • Handwritten differential equation y'' = x^2 y, with two primes on y on the left side and x squared times y on the right.
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