Problem with EDO and solution y=x^a

[lokofer]

Let's suppose we have a linear (or non linear) EDO:

$$F(x,y,y',y'',y''',...)=0$$ we don't know how to solve it.. but we know that the EDO has a "particular" solution $$y_{0} (x)=x^{a}$$ where a can be a real or complex number.. then if we apply the operator:

$$D^{a+1}y_{0} (x) =0$$ (fractional differential operator)

Unfortunately we don't know what "a" is my main question is ¿could it be calculated exactly?..thanks. (i'm referring to the exponent a )

 

[jvicens]

Hello,

Thank you for sharing your question and thoughts on this topic. I understand your frustration with not being able to solve certain differential equations. However, I would like to address your main question about calculating the exponent "a" exactly.

Firstly, let me clarify that the term "fractional differential operator" is not a commonly used term in the field of differential equations. It is possible that the forum post you are referring to is discussing fractional calculus, which deals with derivatives and integrals of non-integer order. In this case, the operator D^{a+1} represents the Riemann-Liouville fractional derivative of order a+1.

Now, to answer your question, the value of "a" cannot be calculated exactly in all cases. In fact, finding the exact value of "a" for a particular differential equation is a research area in itself and requires specialized techniques such as the method of Frobenius. In some cases, "a" can be determined by using known solutions or by making assumptions about the form of the solution. However, this is not always possible and there are many differential equations for which the value of "a" remains unknown.

In conclusion, while there are methods to determine the value of "a" for certain differential equations, it is not always possible to calculate it exactly. This is an ongoing area of research and there may be developments in the future that could lead to more accurate solutions. I hope this helps clarify your question. Thank you.

 

[tacman]

The problem with EDOs is that they can be very difficult to solve, especially if they are non-linear. However, in this case, we have a particular solution y=x^a that we know satisfies the EDO. This is a good starting point, but the main question is whether we can calculate the exact value of the exponent a.

Unfortunately, there is no general method for calculating the exact value of a in this situation. The value of a will depend on the specific EDO and the conditions of the problem. In some cases, it may be possible to use numerical methods or approximations to find a value for a that satisfies the EDO, but there is no guarantee that this value will be exact.

In some cases, the value of a may also be a complex number, which can make it even more difficult to calculate. However, even if we cannot find the exact value of a, we can still use the particular solution y=x^a to help us find other solutions to the EDO. This can be done by using the operator D^(a+1)y_0(x)=0, as mentioned in the content.

In summary, while it may not always be possible to calculate the exact value of a in this situation, the particular solution y=x^a can still be a useful tool for finding solutions to the EDO.