Partial differential equations , rearranging and spotting

In summary: R/dr^2) + 2r(dR/dr) - lambda*R = 0It says to assume R~ r^βThen i can't seem to spot how from that information we can produce this equation:β(β − 1) rβ + 2β rβ − λ rβ = 0Any help would be appreciated, thanks.In summary, the equation given is a differential equation and by assuming R~r^β, we can substitute it into the equation and simplify it to become β(β − 1) rβ + 2β rβ − λ rβ = 0. This can be further simplified by using the substitution t
  • #1
mohsin031211
9
0
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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  • #2
mohsin031211 said:
I have been given an equation : r^2 ( d^2R/dr^2) + 2r(dR/dr) - lambda*R = 0It 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 = 0Any help would be appreciated, thanks.
Simply substitute [itex]R=r^\beta[/itex] into the differential equation.
 
  • #3
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].
 
  • #4
mohsin031211 said:
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 )
 
  • #5
stallionx said:
Set R= constant * r**beta so dR=constant * whatever

They are proportional
R=k*r--beta , all k's ( constants cancel )
Why the constant?
 
  • #6
Hootenanny said:
Why the constant?

Well because I thought tilde is for (constant) linear proportionality.
 
  • #7
I have been given an equation
 

1. What are partial differential equations?

Partial differential equations (PDEs) are mathematical equations that involve multiple variables and their partial derivatives. They are used to describe physical phenomena such as heat transfer, fluid dynamics, and quantum mechanics.

2. How do you rearrange a partial differential equation?

To rearrange a PDE, you can use algebraic operations to isolate the variable you are interested in. Make sure to keep track of the derivatives and their corresponding variables.

3. What does "spotting" mean in the context of partial differential equations?

In the context of PDEs, "spotting" refers to the process of identifying the type of PDE based on its coefficients and variables. This is important for choosing the appropriate solution method for the equation.

4. What are some common methods for solving partial differential equations?

Some common methods for solving PDEs include separation of variables, method of characteristics, and finite difference methods. The choice of method depends on the type of PDE and the boundary conditions.

5. How are partial differential equations used in real-world applications?

PDEs are used in a wide range of real-world applications, including engineering, physics, economics, and biology. They are essential for modeling and understanding complex systems and phenomena, and are often solved using computer simulations.

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