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**Thanksgiving Bonus / Re: Thanksgiving bonus 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

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

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

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

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

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