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MAT244--2018F => MAT244--Tests => Thanksgiving Bonus => Topic started by: Victor Ivrii on October 05, 2018, 05:31:47 PM

Title: Thanksgiving bonus 1
Post by: Victor Ivrii on October 05, 2018, 05:31:47 PM
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
Title: Re: Thanksgiving bonus 1
Post by: Pengyun Li on October 05, 2018, 08:52:24 PM
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: $(-x^4+2x^2-1)y^{''}-(6x^3+2x)y^{'}+(-4x^2)y = 0$.
Title: Re: Thanksgiving bonus 1
Post by: Yiwei Han on October 05, 2018, 09:12:44 PM
but this question for w in the 2nd row is y2''...
Title: Re: Thanksgiving bonus 1
Post by: Jiacheng Ge on October 05, 2018, 09:18:48 PM
Happy Thanksgiving.
Title: Re: Thanksgiving bonus 1
Post by: Jiacheng Ge on October 05, 2018, 10:17:40 PM
Title: Re: Thanksgiving bonus 1
Post by: Jiacheng Ge on October 05, 2018, 10:27:37 PM
By the way, I think my answer to the original question is correct.
Title: Re: Thanksgiving bonus 1
Post by: Pengyun Li on October 05, 2018, 10:37:55 PM
Hi Sir, I figured out the problem and corrected the answer below! Thank you!
Title: Re: Thanksgiving bonus 1
Post by: Monika Dydynski on October 05, 2018, 10:55:40 PM
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 ${2x\over(x-1)^4}$
Title: Re: Thanksgiving bonus 1
Post by: Pengyun Li on October 05, 2018, 11:15:44 PM
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: $2y+4xy'+(x^2-1)y'' = 0$.