Finding the value of a variable present in two functions

$$kx-3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{2}+k$$

$$kx-3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{2}+k$$

honigtropfenvi
2022-09-17
Answered

Finding the value of a variable present in two functions

$$kx-3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{2}+k$$

$$kx-3\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}{x}^{2}+k$$

You can still ask an expert for help

asked 2022-07-09

On a system of PDE

Let ${f}_{1}({p}_{1},{p}_{2})$ and ${f}_{2}({p}_{1},{p}_{2})$ be two functions. The system of equations on ${\mathbb{R}}^{2}$ is just:

${p}_{1}{\mathrm{\partial}}_{{p}_{1}}{f}_{1}+{p}_{2}{\mathrm{\partial}}_{{p}_{1}}{f}_{2}=0$

${p}_{1}{\mathrm{\partial}}_{{p}_{2}}{f}_{1}+{p}_{2}{\mathrm{\partial}}_{{p}_{2}}{f}_{2}=0$

So nobody complain, I want the solutions to be at least of class ${C}^{2}$.

Let ${f}_{1}({p}_{1},{p}_{2})$ and ${f}_{2}({p}_{1},{p}_{2})$ be two functions. The system of equations on ${\mathbb{R}}^{2}$ is just:

${p}_{1}{\mathrm{\partial}}_{{p}_{1}}{f}_{1}+{p}_{2}{\mathrm{\partial}}_{{p}_{1}}{f}_{2}=0$

${p}_{1}{\mathrm{\partial}}_{{p}_{2}}{f}_{1}+{p}_{2}{\mathrm{\partial}}_{{p}_{2}}{f}_{2}=0$

So nobody complain, I want the solutions to be at least of class ${C}^{2}$.

asked 2022-07-01

Find the solutions of this linear equation:

$\begin{array}{}\text{(}\star \text{)}& X+({X}^{\mathrm{\top}}C)C=D\end{array}$

where $C=({c}_{1},\dots ,{c}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ and $D=({d}_{1},\dots ,{d}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ are given and non-zero.

I noticed that the equation $(\star )$ is equivalent to:

$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$

where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}(\mathbb{R})$ such that: $\stackrel{~}{C}}_{i,j}=\{\begin{array}{ll}{c}_{j}{c}_{i}& \text{if}i\ne j\\ 1+{c}_{i}^{2}& \text{if}i=j\end{array$. To do so, I just said that the $i$-th component of the LHS of $(\star )$ is $(1+{c}_{i}^{2}){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$

Is there another method to prove that $(\star )$ is equivalent to $(\star \star )$?

$\begin{array}{}\text{(}\star \text{)}& X+({X}^{\mathrm{\top}}C)C=D\end{array}$

where $C=({c}_{1},\dots ,{c}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ and $D=({d}_{1},\dots ,{d}_{n}{)}^{\mathrm{\top}}\in {\mathbb{R}}^{n}$ are given and non-zero.

I noticed that the equation $(\star )$ is equivalent to:

$\begin{array}{}\text{(}\star \star \text{)}& \stackrel{~}{C}X=D\end{array}$

where $\stackrel{~}{C}\in {\mathrm{M}\mathrm{a}\mathrm{t}}_{n}(\mathbb{R})$ such that: $\stackrel{~}{C}}_{i,j}=\{\begin{array}{ll}{c}_{j}{c}_{i}& \text{if}i\ne j\\ 1+{c}_{i}^{2}& \text{if}i=j\end{array$. To do so, I just said that the $i$-th component of the LHS of $(\star )$ is $(1+{c}_{i}^{2}){x}_{i}+\sum _{j\ne i}{x}_{j}{c}_{j}{c}_{i}$

Is there another method to prove that $(\star )$ is equivalent to $(\star \star )$?

asked 2021-09-04

Solve the system of equations by hand.

asked 2022-09-11

Transformation of inverse to a system of linear equations. Need to solve $X=({U}^{\prime}WU{)}^{-1}{U}^{\prime}$. ${U}^{\prime}$ is ${U}^{\prime}$ is $7\times 7$ positive definite matrix, ${U}^{\prime}$ is of rank $$3$$.

Transformed $({U}^{\prime}WU{)}^{-1}{U}^{\prime}$ as

$({U}^{\prime}WU{)}^{-1}{U}^{\prime}WU=I\phantom{\rule{0ex}{0ex}}XWU=I\phantom{\rule{0ex}{0ex}}{U}^{\prime}W{X}^{\prime}=I\phantom{\rule{0ex}{0ex}}(I\otimes {U}^{\prime}W)vec({X}^{\prime})=vec(I).\phantom{\rule{0ex}{0ex}}$

When I solved $X=({U}^{\prime}WU{)}^{-1}{U}^{\prime}$ and as the above linear system using $$R$$, the answers are slightly different. Does something wrong with the above logic?

Transformed $({U}^{\prime}WU{)}^{-1}{U}^{\prime}$ as

$({U}^{\prime}WU{)}^{-1}{U}^{\prime}WU=I\phantom{\rule{0ex}{0ex}}XWU=I\phantom{\rule{0ex}{0ex}}{U}^{\prime}W{X}^{\prime}=I\phantom{\rule{0ex}{0ex}}(I\otimes {U}^{\prime}W)vec({X}^{\prime})=vec(I).\phantom{\rule{0ex}{0ex}}$

When I solved $X=({U}^{\prime}WU{)}^{-1}{U}^{\prime}$ and as the above linear system using $$R$$, the answers are slightly different. Does something wrong with the above logic?

asked 2022-05-19

Determine the value of $b$ for which the system

$\begin{array}{rl}{x}_{1}+4{x}_{2}-3{x}_{3}+2{x}_{4}& =2\\ 2{x}_{1}+7{x}_{2}-4{x}_{3}+4{x}_{4}& =3\\ -{x}_{1}-5{x}_{2}+5{x}_{3}-2{x}_{4}& =b\\ 3{x}_{1}+10{x}_{2}-5{x}_{3}+6{x}_{4}& =4\end{array}$

is soluble, and determine the solution set.

$\begin{array}{rl}{x}_{1}+4{x}_{2}-3{x}_{3}+2{x}_{4}& =2\\ 2{x}_{1}+7{x}_{2}-4{x}_{3}+4{x}_{4}& =3\\ -{x}_{1}-5{x}_{2}+5{x}_{3}-2{x}_{4}& =b\\ 3{x}_{1}+10{x}_{2}-5{x}_{3}+6{x}_{4}& =4\end{array}$

is soluble, and determine the solution set.

asked 2022-07-13

For the system

$\{\begin{array}{rcrcrcr}x& +& 3y& -& z& =& -4\\ 4x& -& y& +& 2z& =& 3\\ 2x& -& y& -& 3z& =& 1\end{array}$

what is the condition to determine if there is no solution or unique solution or infinite solution?

$\{\begin{array}{rcrcrcr}x& +& 3y& -& z& =& -4\\ 4x& -& y& +& 2z& =& 3\\ 2x& -& y& -& 3z& =& 1\end{array}$

what is the condition to determine if there is no solution or unique solution or infinite solution?

asked 2022-06-29

Explain method to solve two equations in two unknowns where one of the variables has a square term

Given:

$-\frac{96}{{x}^{2}y}+1+y=0$

$-\frac{96}{x{y}^{2}}+2+x=0$

Solve for $x$ and $y$

How should I find $x$ and $y$?

Given:

$-\frac{96}{{x}^{2}y}+1+y=0$

$-\frac{96}{x{y}^{2}}+2+x=0$

Solve for $x$ and $y$

How should I find $x$ and $y$?