$\begin{array}{rl}\dot{x}& =2000-3xy-2x\\ \dot{y}& =3xy-6y\\ \dot{z}& =4y-2z\end{array}$

racodelitusmn
2022-07-03
Answered

Solve the following system of differential equations

$\begin{array}{rl}\dot{x}& =2000-3xy-2x\\ \dot{y}& =3xy-6y\\ \dot{z}& =4y-2z\end{array}$

$\begin{array}{rl}\dot{x}& =2000-3xy-2x\\ \dot{y}& =3xy-6y\\ \dot{z}& =4y-2z\end{array}$

You can still ask an expert for help

asked 2022-06-27

If ${2}^{x}={3}^{y}={6}^{-z}$ and $x,y,z\ne 0$ then prove that:

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

I have tried starting with taking logartithms, but that gives just some more equations.

Any specific way to solve these type of problems?

$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0$

I have tried starting with taking logartithms, but that gives just some more equations.

Any specific way to solve these type of problems?

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$({a}_{11}x+{a}_{12}y)\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}8={b}_{1}$

$({a}_{21}x+{a}_{22}y)\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}8={b}_{2}$

In the case of simple equations without modulo, the solution of $Ax=b$ is $x={A}^{-1}b$, but how to handle this modulo?

$({a}_{21}x+{a}_{22}y)\phantom{\rule{1em}{0ex}}\mathrm{mod}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}8={b}_{2}$

In the case of simple equations without modulo, the solution of $Ax=b$ is $x={A}^{-1}b$, but how to handle this modulo?

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Solve the matrix equation: $4[2x,y,3z]+3[2,-4,6]=[20,-4,54]$

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What are the solutions to the following system of equations?

asked 2020-11-08

4x−6y=12

−2x+3y=−6

−2x+3y=−6

asked 2022-07-09

Is it wrong to write a linear system as below?

Suppose we have the following linear system

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 2{x}_{2}-3{x}_{3}& =6\\ 0& =0.\end{array}$

Is it wrong to write the above linear system by including all the zero coefficients as below

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 0{x}_{1}+2{x}_{2}-3{x}_{3}& =6\\ 0{x}_{1}+0{x}_{2}+0{x}_{3}& =0\end{array}$

instead?

Suppose we have the following linear system

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 2{x}_{2}-3{x}_{3}& =6\\ 0& =0.\end{array}$

Is it wrong to write the above linear system by including all the zero coefficients as below

$\begin{array}{rl}2{x}_{1}+3{x}_{2}+4{x}_{3}& =1\\ 0{x}_{1}+2{x}_{2}-3{x}_{3}& =6\\ 0{x}_{1}+0{x}_{2}+0{x}_{3}& =0\end{array}$

instead?

asked 2021-09-04

Solve the system of equations by hand.