### Author Topic: Thanksgiving bonus 3  (Read 4690 times)

#### Victor Ivrii

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##### Thanksgiving bonus 3
« on: October 05, 2018, 05:34:01 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 $\{y_1(x),y_2(x)\}=\displaystyle{\{x^3,e^x\}}$.
« Last Edit: October 05, 2018, 06:45:43 PM by Victor Ivrii »

#### Tzu-Ching Yen

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##### Re: Thanksgiving bonus 3
« Reply #1 on: October 05, 2018, 06:35:26 PM »
$y_1 = x^3, y_1' = 3x^2, y_1'' = 6x$
$y_2 = y_2' = y_2'' = e^x$
Expand W(y, y_1, y_2)
$$y(y_1'y_2'' - y_2'y_1'') - y'(y_1y_2'' - y_2y_1'') + y''(y_1y_2' - y_2y_1') = 0.$$
Plug in $y_1$, $y_2$ and their derivatives, equation becomes
$$(x^3 - 3x^2)e^xy'' + (6x - x^3)e^xy' + (3x^2 - 6x)e^xy' = 0$$
or
$$(x^2 - 3x)y'' + (6 - x^2)y' + (3x - 6)y' = 0$$

Btw, I think typo is still there. Two $y_2''$.  FIXED
« Last Edit: October 05, 2018, 07:16:03 PM by Victor Ivrii »

#### Jiexuan Wei

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##### Re: Thanksgiving bonus 3
« Reply #2 on: October 05, 2018, 06:44:50 PM »
Here is my answer.  Happy Thanksgiving

#### Monika Dydynski

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##### Re: Thanksgiving bonus 3
« Reply #3 on: October 05, 2018, 07:12:37 PM »
Find a second order equation with the fundamental system of solutions $\lbrace y_1(x),y_2(x) \rbrace =\lbrace x^3,e^x\rbrace$

$$W(x^3,e^x)=\left|\begin{matrix}x^3&e^x\\ 3x^2&e^x\end{matrix}\right|=x^3e^x-3x^2e^x\ne0$$

$$W(y,x^3,e^x)=\left|\begin{matrix}y&x^3&e^x\\ y'&3x^2&e^x\\y''&6x&e^x\end{matrix}\right|=y(3x^2e^x-6xe^x)-y'(x^3e^x-6xe^x)+y''(x^3e^x-3x^2e^x)=0$$

Since $x^3e^x-3x^2e^x\ne0$,

$$y''-{(x^3e^x-6xe^x) \over (x^3e^x-3x^2e^x)}y'+{(3x^2e^x-6xe^x) \over (x^3e^x-3x^2e^x)}y=0$$

Thus, a second order equation with the fundamental system of solutions $\lbrace y_1(x),y_2(x) \rbrace =\lbrace x^3,e^x\rbrace$ is

$$y''-{(x^2-6) \over (x^2-3x)}y'+{(3x-6) \over (x^2-3x)}y=0$$