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Victor Ivrii:
If we want to find a second order equations with the fundamental system of solutions $\{y_1(x),y_2(x)\}$ s.t. $W(y_1,y_2):=\left|\begin{matrix} y_1 & y_2\\ y_1' &y_2'\end{matrix}\right|\ne 0$, we write
$$W(y,y_1,y_2):=\left|\begin{matrix}y & y_1 & y_2\\ y' &y_1' &y_2'\\ y'' &y_1''&y_2'' \end{matrix}\right|= 0.$$

Problem.
Find a second order equation with the fundamental system of solutions $\{y_1(x),y_2(x)\}=\displaystyle{\{\frac{1}{x+1},\frac{1}{x-1}\}}$.

The error corrected

Pengyun Li:
We want to find a second order equation with the fundamental system of solutions $\{y_1(x),y_2(x)\} = \{\frac{1}{x+1}, \frac{1}{x-1}\}$.

Then $y_1=\frac{1}{x+1}, y_2=\frac{1}{x-1}$.

$\implies$ $W(y,y_1,y_2) = W(y, \frac{1}{x+1},\frac{1}{x-1})$ = $\left|\begin{matrix}y & \frac{1}{x+1} & \frac{1}{x-1}\\ y' &-\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\\ y'' &\frac{2x}{(x+1)^4}& \frac{2x}{(x-1)^4} \end{matrix}\right|= 0.$

$\implies$ Solve the determinant : $y\ \left|\begin{matrix} -\frac{1}{(x+1)^2} & -\frac{1}{(x-1)^2}\\ \frac{2x}{(x+1)^4} &\frac{2x}{(x-1)^4}\end{matrix}\right| - y^{'}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ \frac{2x}{(x+1)^4} &\frac{2x}{(x-1)^4}\end{matrix}\right| + y^{''}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ -\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\end{matrix}\right|$

= $y\ (-\frac{8x^2}{(x+1)^4(x-1)^4}) - y^{'}\frac{12x^3+4x}{(x+1)^4(x-1)^4} + y^{''}\frac{-2}{(x+1)^2(x-1)^2} = 0$

Wrong second derivatives!

Then we multiply both sides with $(x-1)^4(x+1)^4$ to get:

$y(-8x^2) - y^{'}(12x^3+4x) + y^{''}(-2x^4+4x^2-2) = 0$

our final equation is: $(-x^4+2x^2-1)y^{''}-(6x^3+2x)y^{'}+(-4x^2)y = 0$.

Yiwei Han:
but this question for w in the 2nd row is y2''...

Jiacheng Ge:
Happy Thanksgiving.

Jiacheng Ge:

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