# Use Wronskian to show only 2 indepndent solutions of 2nd order ODE

• MHB
• ognik
In summary, the author tried various combinations of derivatives but they all led to the same result - the functions could not be linearly independent.
ognik
Given standard ODE $y'' + P(x)y' + Q(x)y=0$, use wronskian to show it cannot have 3 independent sltns. Assume a 3rd solution and show W vanishes for all x.

so 1st row of W = {${y}_{1}, {y}_{2},{y}_{3}$}, 2nd row is 1st derivatives, 3rd row is 2nd derivatives.

I can find the determinate, W (should it be written as |W| ?) = $y_1 y'_2y''_3 - y_1y''_2y'_3 - y_2y'_1y''_3 + y_2y''_1y'_3 + y_3y'_1y''_2 - y_3y''_1y'_2$

I know that $y_1y'_2 - y_2y'_1 \ne 0$(IE $y_3$ is the 'other' solution), so I grouped the above into $y''_3(y_1y'_2 - y_2y'_1) - y''_2(y_1y'_3 - y_3y'_1) - y''_1(y_2y'_3 - y_3y'_1)$. The terms in brackets are Wronskians and if the 3 solutions are linearly independent, then they cannot=0. It would be convenient then, to have the 2nd derivatives = 0, but I can't argue that as the y(x)'s could have $x^2$ terms...

I have tried various other combinations, but most of them just transform to expansions by a different row/column. Can anyone give me a hint please?

Assuming your Wronskian is correct, maybe you could substitute $y_2''=-P(x) y_2'-Q(x) y_2$, and the same for the other two solutions?

... and they all cancel neatly, I wish I could think of these obvious-in-retrospect approaches.

Well, in this case, what led me to that suggestion was that, so far, you had not used the fact that the functions were solutions. That had to be important.

I also thought it important, but instead thought that suggested using the Wronskian, I think with a lot more practice it gets better, its just frustrating when missing something obvious costs me many hours ... without this forum I suspect I would have given up by now, so I really appreciate all the help!

## 1. What is the Wronskian method used for in solving second order ODEs?

The Wronskian method is a mathematical tool that is used to determine whether a set of functions are linearly independent or not.

## 2. How does the Wronskian method work?

In the context of solving second order ODEs, the Wronskian method involves calculating the determinant of a matrix composed of the two given solutions of the ODE. If the determinant is non-zero, then the solutions are linearly independent and can be used as a basis for the general solution of the ODE.

## 3. Why is it important to show that only two solutions are independent in a second order ODE?

In most cases, second order ODEs have two independent solutions that can be used to obtain the general solution. Showing that only two solutions are independent ensures that the general solution is not over-determined, which would result in an invalid solution.

## 4. Can the Wronskian method be used for higher order ODEs?

Yes, the Wronskian method can be extended to higher order ODEs by creating a matrix with more rows and columns corresponding to the number of solutions.

## 5. What are the limitations of using the Wronskian method to show independent solutions in second order ODEs?

The Wronskian method can only be used to determine linear independence, and cannot be used to prove that solutions are linearly dependent. It also does not provide an efficient way to obtain the general solution, as it only identifies the independent solutions.

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