Solve the system of equations

Isa Trevino
2021-06-18
Answered

Solve the system of equations

You can still ask an expert for help

Khribechy

Answered 2021-06-19
Author has **100** answers

Given that

and

From (1)

Therefore from (1) we have

Jeffrey Jordon

Answered 2021-10-10
Author has **2262** answers

Answer is given below (on video)

asked 2020-12-28

Solve the system:

asked 2022-06-25

So my course shows me three differential equations:

$\dot{x}+{x}^{2}=t$

$\dot{x}=({t}^{2}+1)(x-1)$

$\dot{x}+x={t}^{2}$

The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.

Unfortunately, they don't show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?

For the second one, I thought I bring it into standard form somehow:

$\dot{x}=({t}^{2}+1)(x-1)=x{t}^{2}+x-{t}^{2}-1=x({t}^{2}+1)-{t}^{2}-1$

If we say we let $p(t)={t}^{2}+1$ and $q(t)=1+{t}^{2}$, then we could say:

$\dot{x}=xp(t)-q(t)=...$

And so on, to simplify until we reach standard form of a linear ODE (or not).

Is that the way to go? Or is there some other way to check if a ODE is linear?

$\dot{x}+{x}^{2}=t$

$\dot{x}=({t}^{2}+1)(x-1)$

$\dot{x}+x={t}^{2}$

The first one is not a linear ordinary differential equation (ODE) apparently, the other two are.

Unfortunately, they don't show a clear way how to find out if an ODE is linear or not. So how we can find out if an ODE is linear?

For the second one, I thought I bring it into standard form somehow:

$\dot{x}=({t}^{2}+1)(x-1)=x{t}^{2}+x-{t}^{2}-1=x({t}^{2}+1)-{t}^{2}-1$

If we say we let $p(t)={t}^{2}+1$ and $q(t)=1+{t}^{2}$, then we could say:

$\dot{x}=xp(t)-q(t)=...$

And so on, to simplify until we reach standard form of a linear ODE (or not).

Is that the way to go? Or is there some other way to check if a ODE is linear?

asked 2021-12-11

Find the absolute value.

|1-i|

|1-i|

asked 2021-09-06

If $\mathrm{sin}\left(\theta \right)=-\frac{1}{4}\mathrm{sin}$ and $\theta$ is in the 3rd quadrant, find $\mathrm{cos}\left(\theta \right)$

asked 2021-12-07

Solve.

$\left|\frac{x-7}{x+5}\right|\ge -5$

asked 2022-06-29

My problem:

${x}_{1}+3{x}_{2}-{x}_{3}+{x}_{4}+2{x}_{5}=2,$

$2{x}_{1}-{x}_{2}+{x}_{3}-{x}_{4}+{x}_{5}=-2,$

$4{x}_{1}+2{x}_{2}+2{x}_{3}-{x}_{4}-{x}_{5}=0.$

I thought that I could multiply 1 by equation 2, and then add that to equation 1, which would cancel both the ${x}_{3}$ and the ${x}_{4}$ factors for that line. Is that permissible?

${x}_{1}+3{x}_{2}-{x}_{3}+{x}_{4}+2{x}_{5}=2,$

$2{x}_{1}-{x}_{2}+{x}_{3}-{x}_{4}+{x}_{5}=-2,$

$4{x}_{1}+2{x}_{2}+2{x}_{3}-{x}_{4}-{x}_{5}=0.$

I thought that I could multiply 1 by equation 2, and then add that to equation 1, which would cancel both the ${x}_{3}$ and the ${x}_{4}$ factors for that line. Is that permissible?

asked 2022-04-27

Solve the inequality

$\frac{-5x{(x-3)}^{2}}{2+x}\le 0$