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Question

# Transform the equation ${t(t - 1)y^{(4)} + e^{t}y'' + 4t^{2}y = 0}$ into an equivalent first order system.

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We have the equation

$\begin{gather*} t(t - 1)y^{(4)} + e^{t} y'' + 4t^{2} y = 0, \end{gather*}$

and we have to transform it into an equivalent first order system by the following technique :

Let the functions

$\begin{gather*} x_{1} = y, \\ x_{2} = y', \\ x_{3} = y'', \\ x_{4} = y''', \end{gather*}$

and differentiate them with respect to $t$,

Then we have

$\begin{gather*} x'_{1} = y' = x_{2}, \tag{1} \\ x'_{2} = y'' = x_{3}, \tag{2} \\ x'_{3} = y''' = x_{4}, \tag{3} \\ x'_{4} = y^{(4)} = - \dfrac{e^{t}}{t(t - 1)} y'' - \dfrac{4t}{t - 1} y = - \dfrac{e^{t}}{t(t - 1)} x_{3} - \dfrac{4t}{t - 1} x_{1} \tag{4} \\ \end{gather*}$

Put equations (1), (2), (3) and (4) into a matrix type, then we have

$\begin{gather*} \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3}\end{pmatrix}' = \begin{pmatrix}0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ -\dfrac{4t}{t - 1} & 0 & -\dfrac{e^{t}}{t(t - 1)} & 0\end{pmatrix} \ \begin{pmatrix}x_{1} \\ x_{2} \\ x_{3} \\ x_{3}\end{pmatrix} + \begin{pmatrix}0 \\ 0 \\ 0 \\ 0\end{pmatrix} \end{gather*}$

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