Why Must B Approach Zero in Cauchy-Euler ODE Solutions at x=0?

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In summary, the conversation discusses the application of boundary conditions to ODEs and their solutions. It is stated that when certain boundary conditions, such as x=0, are applied, the solution takes the form of Ax^k + Bx^(-k), where B must go to zero. This is due to the fact that the negative exponent of the B coefficient term implies a divergent solution at x=0, which contradicts the physical interpretation of solutions such as electron wave functions. It is then explained that setting B=0 resolves this issue. The conversation ends with the realization that this solution may seem simple, but it effectively resolves any confusion or existential questions.
  • #1
Andreol263
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Why these ODEs when applied some boundary conditions, like x = 0, their solution of the form Ax^k + Bx^(-k), B WILL have to go to zero?Like some problems which involve spherical harmonics...
 
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  • #2
The negative exponent of the B coefficient term implies divergent solution at x=0. If you are considering solutions with physical interpretation such as electron wave functions then the physical application is contradicted by the divergence at infinity. Those solutions don't apply and you set B = 0.
 
  • #3
Yeah i understand why when x->infinity the solution is inconsistent, cause it's blow up, but i simply don't get it WHY x=0 implies in a divergent solution and B has to be 0, there's a more deep explanation?
 
  • #4
[itex]B x^{-k}=\frac{B}{x^k} = \frac{B}{0}[/itex] when [itex]x=0[/itex].
 
  • #5
Oh damn, I'm stupid , i was thinking in that, but it appears so simple that i forget this option because some texts are very confusing , and thank you so much for killing that existential question for me :D...
 

Related to Why Must B Approach Zero in Cauchy-Euler ODE Solutions at x=0?

Question 1: What is a Cauchy-Euler ODE?

An ODE (ordinary differential equation) of the form axny(n) + bx(n-1)y(n-1) + ... + cx'y + dy = 0, where a, b, c, and d are constants and n is a non-negative integer.

Question 2: How do you solve a Cauchy-Euler ODE?

To solve a Cauchy-Euler ODE, you can use the substitution y = xm. This will transform the ODE into a polynomial equation, which can then be solved using methods such as factoring or the quadratic formula.

Question 3: What is the characteristic equation in a Cauchy-Euler ODE?

The characteristic equation in a Cauchy-Euler ODE is the polynomial equation that results from substituting y = xm into the ODE. It is used to find the values of m that will yield a solution to the ODE.

Question 4: Can a Cauchy-Euler ODE have non-integer exponents?

Yes, a Cauchy-Euler ODE can have non-integer exponents. In this case, the substitution y = xm is still used, but the characteristic equation will result in complex roots, which will yield complex solutions for y.

Question 5: What are some real-world applications of Cauchy-Euler ODEs?

Cauchy-Euler ODEs can be used to model various physical phenomena, such as the motion of a pendulum, the decay of radioactive materials, and the flow of fluids. They are also commonly used in engineering and economics to model systems with changing variables over time.

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