No Analytic Solution for ODE x^3y'-2y+2x=0

[Malmstrom]

Homework Statement

Consider the following ODE
$$x^3y' - 2y + 2x = 0$$

Homework Equations

Prove that the ODE has no analytic solution in any neighborhood of $$x=0$$.

The Attempt at a Solution

Its general solution is $$Ce^{-\frac{1}{x^2}}+ e^{-\\frac{1}{x^2}} \int_{x_0}^x - \frac{2e^{\frac{1}{t^2}}}{t^2} dt$$ which is not continuous in $$x=0$$, but I don't think this is unique so I don't know if it helps. Should I try to substitute an arbitrary power series in the equation?

 

[mighty2000]

you should approach this problem using mathematical rigor and logical reasoning. First, let's define what an analytic solution is. An analytic solution is a solution that can be expressed as a power series, where each term in the series has a finite number of derivatives at a given point.

Now, let's consider the given ODE: x^3y' - 2y + 2x = 0. If we try to solve this using the method of separation of variables, we get the following solution: y(x) = Ce^(1/x^2) + e^(1/x^2) ∫(x_0)^x - (2e^(1/t^2))/(t^2) dt. However, if we try to evaluate this solution at x=0, we get an undefined value. This means that the solution is not continuous at x=0, which contradicts the definition of an analytic solution.

To further prove that there is no analytic solution in any neighborhood of x=0, we can also use the Picard-Lindelof theorem. This theorem states that if a function and its partial derivative are continuous in a given region, then there exists a unique solution to the ODE in that region. However, in this case, the partial derivative of the solution is not continuous at x=0, which means that there is no unique solution in any neighborhood of x=0.

Therefore, we can conclude that the given ODE has no analytic solution in any neighborhood of x=0.