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We suppose that we have a system of differential equations $$\left\{\begin{matrix} L_1 y=0\\ L_2 y=0 \end{matrix}\right. \tag 1$$ where there order of $L_1 y$ is $n$ and the order of $L_2 y$ is $m$ and the coefficients of the highest-order term is $1$.

Let $n<m$.

Let $d_jL_i$ be the $j$th derivative of $L_i y$.

We define the differential equation $d_{m-n}L_1 y-L_2 y=0$ the order of which is smaller than the order of $L_2 y$.

So the system $(1)$ is equivalent to the system $$\left\{\begin{matrix} L_1 y=0\\ d_{m-n}L_1 y-L_2 y=0 \end{matrix}\right.$$

So the initial system is equivalent to a system of smaller order.

We do the same procedure until we get a system of the form $$\left\{\begin{matrix} L y=0\\ \alpha \end{matrix}\right.$$ where the order of $L y$ is $0$ and $\alpha$ doesn't contain $x$, i.e., it is a relation between the parameters.

So we conclude to one differential equation.

That means that any system can be replaced by one differential equation.

Is this correct?

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Example:

We have the system :

$$\left\{\begin{matrix} L_1 y=y'+y=0\\ L_2 y=y''+y'+y=0 \end{matrix}\right.$$

The order of $L_2 y$, which is $2$, is greater than the order of $L_1 y$, which is $1$.

So it is equivalent to the system

$$\left\{\begin{matrix} L_1 y=0\\ L_2' y=d_1L_1 y-L_2 y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1 y=y'+y=0\\ L_2' y=y''+y'-y''-y'-y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1 y=y'+y=0\\ L_2' y=-y=0 \end{matrix}\right.$$

So multiply at the second equation by $-1$ so that the coefficient of the highest order term is $1$.

So we get $$\left\{\begin{matrix} L_1 y=y'+y=0\\ L_2' y=y=0 \end{matrix}\right.$$

The order of $L_1 y$, which is $1$, is greater than the order of $L_2' y$, which is $0$.

So it is equivalent to the system

$$\left\{\begin{matrix} d_1L_2' y-L_1 y=0\\ L_2' y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1' y=y'-y'-y=0\\ L_2' y=y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1' y=-y=0\\ L_2' y=y=0 \end{matrix}\right. \Rightarrow y=0$$

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  • $\begingroup$ can you provide an example? $\endgroup$ – Michael Medvinsky Nov 27 '15 at 20:29
  • $\begingroup$ you also want $Lu=0$ not $L=0$ $\endgroup$ – Michael Medvinsky Nov 27 '15 at 20:30
  • $\begingroup$ I changed $L=0$ to $Ly=0$. I also added an example... @MichaelMedvinsky $\endgroup$ – Mary Star Nov 27 '15 at 20:37
  • $\begingroup$ do the same change in example:)....reading now $\endgroup$ – Michael Medvinsky Nov 27 '15 at 20:39
  • $\begingroup$ I did it... @MichaelMedvinsky $\endgroup$ – Mary Star Nov 27 '15 at 20:42
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Taking another example $$\left\{\begin{matrix} L_1y=ay'+by=0\\ L_2y=cy''+dy'+ey=0 \end{matrix}\right.$$ now this becomre $$\left\{\begin{matrix} L_1y=ay'+by=0\\ L_2'y=ay''+by' + cy''+dy'+ey=(a+ c)y''+(b +d)y'+ey=0 \end{matrix}\right.$$ Note that in this case $L'$ is doesn't really "better" then $L$.

When you talking about linear system of differential equations you should think about vector $y=(x,x',x'',x''',\dots,x^{(n)})^T$ and some matrix of coefficients $A$ and some vector function $b(x)=(b_1(x),\dots,b_n(x))$, such that $$y'=Ay+b$$

Very simple examples

1) $y''+y'+y=0$, denote $v=y'$, $w=y$ then $v'+v+w=0$ become $$\pmatrix{y'\\y''}=\frac{d}{dt}\pmatrix{w\\v}=\pmatrix{0&1\\-1&-1}\pmatrix{w\\v}=\pmatrix{v\\-v-w} $$

2) The other way(different example)

$$\begin{cases}{x'=x+y\\ y'=x-y}\end{cases} \rightarrow \begin{cases}{x''=x'+y'\\ y'=x-y}\end{cases} \rightarrow x''=x'+x-y$$

This is very simmilar to what you want, but how one can solve it in this form?

The system can be solved of course. Rewrite as $$\frac{d}{dt}\begin{pmatrix}x\\y\end{pmatrix} =\begin{pmatrix}1&1\\1&-1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}$$ Since $$\begin{pmatrix}1&1\\1&-1\end{pmatrix}= \begin{pmatrix} 1-\sqrt{2} & 1 \\ 1+\sqrt{2} & 1 \\ \end{pmatrix}^{-1} \begin{pmatrix} -\sqrt{2} & 0 \\ 0 & \sqrt{2} \\ \end{pmatrix}\begin{pmatrix} 1-\sqrt{2} & 1 \\ 1+\sqrt{2} & 1 \\ \end{pmatrix} $$ one multiply the system by $$\begin{pmatrix} 1-\sqrt{2} & 1 \\ 1+\sqrt{2} & 1 \\ \end{pmatrix}$$ to get

$$\begin{pmatrix}\tilde x\\\tilde y\end{pmatrix}= \begin{pmatrix}\left(1-\sqrt{2}\right) x+y\\ \left(1+\sqrt{2}\right) x+y\end{pmatrix}$$ and $$\frac{d}{dt}\begin{pmatrix}\tilde x\\\tilde y\end{pmatrix} =\begin{pmatrix} -\sqrt{2} & 0 \\ 0 & \sqrt{2} \\ \end{pmatrix}\begin{pmatrix}\tilde x\\\tilde y\end{pmatrix}$$

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  • $\begingroup$ Isn't it as follows? $$\left\{\begin{matrix} L_1y=ay'+by=0\\ L_2y=cy''+dy'+ey=0 \end{matrix}\right.$$ We want that the coefficients of the highest order term is $1$, so we divide the differential equations by $a$ and $c$ respectively. So we get $$\left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\ L_2y=y''+\frac{d}{c}y'+\frac{e}{c}y=0 \end{matrix}\right.$$ $\endgroup$ – Mary Star Nov 27 '15 at 23:40
  • $\begingroup$ $$\left\{\begin{matrix} L_1 y=0\\ L_2' y=d_{1}L_1 y-L_2 y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2' y=y''+\frac{b}{a}y'-(y''+\frac{d}{c}y'+\frac{e}{c}y)=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2 ' y = \left (\frac{b}{a}-\frac{d}{c}\right )y'-\frac{e}{c}y=0 \end{matrix}\right.$$ $\endgroup$ – Mary Star Nov 27 '15 at 23:41
  • $\begingroup$ $$\Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2 ' y = y'-\frac{e}{c\left (\frac{b}{a}-\frac{d}{c}\right )}y=0 \end{matrix}\right. $$ $\endgroup$ – Mary Star Nov 27 '15 at 23:41
  • $\begingroup$ $$ \Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2'' y=d_{0}L_1 y-L_2' y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2'' y=y'+\frac{b}{a}y-\left (y'-\frac{e}{c\left (\frac{b}{a}-\frac{d}{c}\right )}y\right )=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2'' y=\left (\frac{b}{a}+\frac{e}{c\left (\frac{b}{a}-\frac{d}{c}\right )} \right ) y=0 \end{matrix}\right. $$ $\endgroup$ – Mary Star Nov 27 '15 at 23:41
  • $\begingroup$ $$\Rightarrow \left\{\begin{matrix} L_1y=y'+\frac{b}{a}y=0\\\ L_2'' y= y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1' y=d_{1}L_2 y-L_1 y=0 \\ L_2'' y= y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1' y=y'-\left (y'+\frac{b}{a}y\right )=0 \\ L_2'' y= y=0 \end{matrix}\right. \Rightarrow \left\{\begin{matrix} L_1' y=-\frac{b}{a}y=0 \\ L_2'' y= y=0 \end{matrix}\right. \Rightarrow y=0$$ $\endgroup$ – Mary Star Nov 27 '15 at 23:42

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