fythynwyrk0

Answered

2022-07-04

Express the differential equation

${y}^{\u2034}-6{y}^{\u2033}-{y}^{\prime}+6y=0$

as a system of first order equations i.e. a matrix equation of the form

$A(\overrightarrow{x}{)}^{\prime}=0$

where

$\overrightarrow{x}\text{is the vector}\left[\begin{array}{r}{x}_{1}\\ \text{}{x}_{2}\\ \text{}{x}_{3}\end{array}\right].$

${y}^{\u2034}-6{y}^{\u2033}-{y}^{\prime}+6y=0$

as a system of first order equations i.e. a matrix equation of the form

$A(\overrightarrow{x}{)}^{\prime}=0$

where

$\overrightarrow{x}\text{is the vector}\left[\begin{array}{r}{x}_{1}\\ \text{}{x}_{2}\\ \text{}{x}_{3}\end{array}\right].$

Answer & Explanation

Elliott Gilmore

Expert

2022-07-05Added 10 answers

Here how you advance, let y'=z, then we will have the system

${z}^{\u2033}-6{z}^{\prime}-z+6y=0\phantom{\rule{0ex}{0ex}}{y}^{\prime}=z$

Again, put z'=w which results in the system

${w}^{\prime}-6w-z+6y=0\phantom{\rule{0ex}{0ex}}{z}^{\prime}=w\phantom{\rule{0ex}{0ex}}{y}^{\prime}=z$

Arranging the above equation gives

${y}^{\prime}=z\phantom{\rule{0ex}{0ex}}{z}^{\prime}=w\phantom{\rule{0ex}{0ex}}{w}^{\prime}=6w+z-6y.$

Now, I am sure you know how to write this in a matrix form.

${z}^{\u2033}-6{z}^{\prime}-z+6y=0\phantom{\rule{0ex}{0ex}}{y}^{\prime}=z$

Again, put z'=w which results in the system

${w}^{\prime}-6w-z+6y=0\phantom{\rule{0ex}{0ex}}{z}^{\prime}=w\phantom{\rule{0ex}{0ex}}{y}^{\prime}=z$

Arranging the above equation gives

${y}^{\prime}=z\phantom{\rule{0ex}{0ex}}{z}^{\prime}=w\phantom{\rule{0ex}{0ex}}{w}^{\prime}=6w+z-6y.$

Now, I am sure you know how to write this in a matrix form.

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