MAT244--2018F > Thanksgiving Bonus

Thanksgiving bonus 1

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Jiacheng Ge:
By the way, I think my answer to the original question is correct.

Pengyun Li:
Hi Sir, I figured out the problem and corrected the answer below! Thank you!

Monika Dydynski:
Pengyun,

$\left({1 \over {(x+1)}}\right)''=\left(-{1 \over {(x+1)^2}}\right)'\ne{2x\over(x+1)^4}$. Instead, $y_1''(x)={2\over(x+1)^3}$

Similarly,

$y_2''(x)={2\over(x-1)^3}$, not ${2x\over(x-1)^4}$

Pengyun Li:
Corrected: :)

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

$\implies$ $W(y,y_1,y_2) = W(y, \frac{1}{x+1},\frac{1}{x-1})$ = $\left|\begin{matrix}y & \frac{1}{x+1} & \frac{1}{x-1}\\ y' &-\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\\ y'' &\frac{2}{(x+1)^3}& \frac{2}{(x-1)^3} \end{matrix}\right|= 0.$

$\implies$ Solve the determinant : $y\ \left|\begin{matrix} -\frac{1}{(x+1)^2} & -\frac{1}{(x-1)^2}\\ \frac{2}{(x+1)^3} &\frac{2}{(x-1)^3}\end{matrix}\right| - y^{'}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ \frac{2}{(x+1)^3} &\frac{2}{(x-1)^3}\end{matrix}\right| + y^{''}\ \left|\begin{matrix} \frac{1}{x+1} & \frac{1}{x-1}\\ -\frac{1}{(x+1)^2} &-\frac{1}{(x-1)^2}\end{matrix}\right|$

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

Then we multiply both sides with $(x-1)^3(x+1)^3$ to get:

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

Our final equation is: $2y+4xy'+(x^2-1)y'' = 0$.