Mario has been changed

[mario99]

Solve

$\frac{dx}{dt} = 2x + 3y$

$\frac{dy}{dt} = 2x + y$

First I tried the elimination

$\frac{dx}{dt} - 2x = 3y$

$dx - 2x \ dt = 3y \ dt$

$x(d - 2 \ dt) = 3y \ dt$

$x = \frac{3y \ dt}{(d - 2 \ dt)}$

$\frac{dy}{dt} = 2(\frac{3y \ dt}{(d - 2 \ dt)}) + y$

I have difficulties to isolate y in this equation.

My second thought is to solve the system by Matrix. I think it is also possible.


Any help would be appreciated.

 

[topsquark]

Solve

$\frac{dx}{dt} = 2x + 3y$

$\frac{dy}{dt} = 2x + y$

First I tried the elimination

$\frac{dx}{dt} - 2x = 3y$

$dx - 2x \ dt = 3y \ dt$

$x(d - 2 \ dt) = 3y \ dt$

$x = \frac{3y \ dt}{(d - 2 \ dt)}$

$\frac{dy}{dt} = 2(\frac{3y \ dt}{(d - 2 \ dt)}) + y$

I have difficulties to isolate y in this equation.

My second thought is to solve the system by Matrix. I think it is also possible.


Any help would be appreciated.

I've gotten tired of the number of times you've told me you've changed. I presume, as always, that you expect all of us to be amazed and forgive everything that has come before. I'll simply take it to mean that you are going to follow the forum rules for at least a short amount of time and that, at some point, you'll revert. You always do.

You supposedly have taken Differential Equations and you tried to factor the x out of dx - 2x dt? dx is not d multiplied by x... the "d" is an operator, not a number. I told you a long time ago that you weren't ready for this level, and this line just proves it.

Okay, a Linear Algebra approach is the best way to handle this, but I don't know how much, if any of that, you've covered. Here's a straightforward, but not elegant, method that you should be able to follow, based solely on Differential Equations.

Take the time derivative of the first equation:
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \dfrac{dy}{dt}$

The second equation gives an expression for dy/dt, so plug that in:
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 ( 2x + y )$

Now, solve the first equation for y and sub that in:
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2x + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2x \right ) \right )$

Now the y has been taken out. Solve for x and plug that into the first original equation and solve for y.

-Dan

 

[Steven G]

What is d in your work? Is it a variable like x and y? Is it an operator?
Also you start off with saying solve! Solve for what? You solved for dy/dt--why did you choose to solve for that?
Personally I would have solved for dx/dt or dy/dt as that was given! You chose to solve for dy/dt. Why not stop at dy/dt = 2x+y?
I too suspect that you are not ready for problems of this level.

 

[mario99]

Thank you topsquark and Steven G for helping me.

Remark to Mr. Dan: Yes I am new Mario now.

I've gotten tired of the number of times you've told me you've changed. I presume, as always, that you expect all of us to be amazed and forgive everything that has come before. I'll simply take it to mean that you are going to follow the forum rules for at least a short amount of time and that, at some point, you'll revert. You always do.

You supposedly have taken Differential Equations and you tried to factor the x out of dx - 2x dt? dx is not d multiplied by x... the "d" is an operator, not a number. I told you a long time ago that you weren't ready for this level, and this line just proves it.

Okay, a Linear Algebra approach is the best way to handle this, but I don't know how much, if any of that, you've covered. Here's a straightforward, but not elegant, method that you should be able to follow, based solely on Differential Equations.

Take the time derivative of the first equation:
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \dfrac{dy}{dt}$

The second equation gives an expression for dy/dt, so plug that in:
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 ( 2x + y )$

Now, solve the first equation for y and sub that in:
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2x + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2x \right ) \right )$

Now the y has been taken out. Solve for x and plug that into the first original equation and solve for y.

-Dan

I don't fully understand your method. I will try to substitute one equation into the other.

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 ( 2x + y )$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 6x + 6y$

$\dfrac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}= x$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}\right] + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}\right] \right ) \right )$

Is this correct?

What is d in your work? Is it a variable like x and y? Is it an operator?
Also you start off with saying solve! Solve for what? You solved for dy/dt--why did you choose to solve for that?
Personally I would have solved for dx/dt or dy/dt as that was given! You chose to solve for dy/dt. Why not stop at dy/dt = 2x+y?
I too suspect that you are not ready for problems of this level.

d is neither variable nor operator, it is just a letter. I am so sure that I did this elimination before and got it right.

What's wrong with you people? If I show no attempts, you complain and when I show attempts you also complain. What do you want to see exactly? Why do you think that I am not ready? I have solved Airy equation from scratch? Can an amateur do that? I am very well prepared for this material, but I have some issues in simplifying.

 

[topsquark]

I don't fully understand your method. I will try to substitute one equation into the other.

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 ( 2x + y )$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 6x + \red{\textbf{3}}y$

(Note a typo that you made above.) The point of the method was to find a way to eliminate y from the first equation. I took the derivative of the first equation
$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \dfrac{dy}{dt}$

subbed in dy/dt from the second equation
$\dfrac{dy}{dt} = 2x + y$

to get rid of that term, then used y from the first equation
$y = \dfrac{1}{3} \left ( \dfrac{dx}{dt} -2x \right )$
to eliminate the y.

d is neither variable nor operator, it is just a letter. I am so sure that I did this elimination before and got it right.

d is not just a letter, it is the differential operator. You can't just treat it like a variable. dx is not at all anything like xd. If you had paid more attention in Calculus you would have that fact solidly ingrained. The fact that you haven't means you didn't spend enough time with the material.

What's wrong with you people? If I show no attempts, you complain and when I show attempts you also complain. What do you want to see exactly? Why do you think that I am not ready? I have solved Airy equation from scratch? Can an amateur do that? I am very well prepared for this material, but I have some issues in simplifying.

I don't really care what you've derived in the past, if you make a mistake that basic you are going to be told about it. I am not complaining that you showed work (I wouldn't have responded at all if you hadn't) I am doing what I can to correct what is a grievous mistake on your part at the level you are insisting on working at. If you did that in Calculus I problem I merely would have mentioned it. I am treating you as if you were a student working at the level of someone who has completed Differential Equations and any such student I have ever taught never would have made that mistake no matter what substance they were high on. I would treat an A level student with the same contempt that I gave you for making that mistake.

If you insist that you understand material at a higher level than you clearly are capable of working at, I'm going to let you know about it.

-Dan

 

[Steven G]

Thank you topsquark and Steven G for helping me.

Remark to Mr. Dan: Yes I am new Mario now.


I don't fully understand your method. I will try to substitute one equation into the other.

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 ( 2x + y )$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 6x + 6y$

$\dfrac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}= x$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}\right] + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}\right] \right ) \right )$

Is this correct?



d is neither variable nor operator, it is just a letter. I am so sure that I did this elimination before and got it right.

What's wrong with you people? If I show no attempts, you complain and when I show attempts you also complain. What do you want to see exactly? Why do you think that I am not ready? I have solved Airy equation from scratch? Can an amateur do that? I am very well prepared for this material, but I have some issues in simplifying.

Why do I think that you are not ready?
Here is why. You think that you are to solve for dy/dt. Fine, no problem. BUT THEY TOLD YOU WHAT DY/DT EQUALS! dy/dt equals 2x+y.
When the answer is given and a student doesn't see it, then I think that the student is not ready for the course that they are in.

For example: 2x + 3y = 7 and 7x-4y =9. If the problems asks for the value for 2x+3y, then I claim this is an amazingly simple problem to answer.

You have two extremely qualified people telling you that you do not yet belong in Differential Equations and you question us. Please try questioning yourself. We both can be wrong with our assessment about you but probably are correct.

 

[khansaheb]

Thank you topsquark and Steven G for helping me.

Remark to Mr. Dan: Yes I am new Mario now.


I don't fully understand your method. I will try to substitute one equation into the other.

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 ( 2x + y )$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 6x + 6y$

$\dfrac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}= x$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}\right] + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{6}\right] \right ) \right )$

Is this correct?



d is neither variable nor operator, it is just a letter. I am so sure that I did this elimination before and got it right.

What's wrong with you people? If I show no attempts, you complain and when I show attempts you also complain. What do you want to see exactly? Why do you think that I am not ready? I have solved Airy equation from scratch? Can an amateur do that? I am very well prepared for this material, but I have some issues in simplifying.

The New Mario,

There are large differences between

Algebraic equation (learnt in Middle/High school) and
Differential equation (DE - learnt in college)

Please investigate those. Please tell us what you find through your text books anb Google.

The equation you posted falls under the category of advanced DE).

Help us help you!!!!!

 

[mario99]

(Note a typo that you made above.)

Yeah I saw it. It happened probably due to speed.

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{3}\right] + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 6y}{3}\right] \right ) \right )$

The point of the method was to find a way to eliminate y from the first equation. I took the derivative

Even If I eliminate y, isn't the main idea to find $x(t)$ and $y(t)$?

I don't see how I will get them from your method.

d is not just a letter, it is the differential operator. You can't just treat it like a variable. dx is not at all anything like xd. If you had paid more attention in Calculus you would have that fact solidly ingrained. The fact that you haven't means you didn't spend enough time with the material.

I don't understand what you are trying to say here, but I am so confident I did the method of elimination before.

I don't really care what you've derived in the past, if you make a mistake that basic you are going to be told about it. I am not complaining that you showed work (I wouldn't have responded at all if you hadn't) I am doing what I can to correct what is a grievous mistake on your part at the level you are insisting on working at. If you did that in Calculus I problem I merely would have mentioned it. I am treating you as if you were a student working at the level of someone who has completed Differential Equations and any such student I have ever taught never would have made that mistake no matter what substance they were high on. I would treat an A level student with the same contempt that I gave you for making that mistake.

I want you to tell me what is the level of a student who is capable to solve the Airy equation?

Why do I think that you are not ready?
Here is why. You think that you are to solve for dy/dt. Fine, no problem. BUT THEY TOLD YOU WHAT DY/DT EQUALS! dy/dt equals 2x+y.
When the answer is given and a student doesn't see it, then I think that the student is not ready for the course that they are in.

For example: 2x + 3y = 7 and 7x-4y =9. If the problems asks for the value for 2x+3y, then I claim this is an amazingly simple problem to answer.

You have extremely qualifies people telling you that you do not yet belong in Differential Equations and you question us. Please try questioning yourself. We both can be wrong with our assessment about you but probably are correct.

You are wrong. The question is not asking for $\frac{dx}{dt}$ or $\frac{dy}{dt}$. It asks to solve the system to find $x(t)$ and $y(t)$. If you are confused or mixed this question with something else, please read about Linear Systems. I am telling that I am 100% qualified for this course. Answer this question please, how was I able to solve the Airy equation if I am not qualified?

The New Mario,

There are large differences between

Algebraic equation (learnt in Middle/High school) and
Differential equation (DE - learnt in college)

Please investigate those. Please tell us what you find through your text books anb Google.

The equation you posted falls under the category of advanced DE).

Help us help you!!!!!

You don't need to tell me the difference between algebraic equation and differential equation. I have solved the Airy equation before and I know exactly what is a differential equation. I have also posted a long problem about partial differential equation whom no one of you was able to solve it. I am not a detective to investigate and if you think this DE belongs to advanced mathematics this means I am an advanced student as well. I can't help you if you can't help me.

 

[topsquark]

I don't see how I will get them from your method.

My method gives you a second order differential equation for x. Solve for x!

I don't understand what you are trying to say here, but I am so confident I did the method of elimination before.

Elimination, as in the algebraic case, will not work here.

I want you to tell me what is the level of a student who is capable to solve the Airy equation?

I want you to show me any competent Differential Equations student who does NOT know that the d in dx is not a variable. Otherwise, do you believe that dx/dt = x/t??

You are wrong. The question is not asking for $\frac{dx}{dt}$ or $\frac{dy}{dt}$. It asks to solve the system to find $x(t)$ and $y(t)$. If you are confused or mixed this question with something else, please read about Linear Systems. I am telling that I am 100% qualified for this course. Answer this question please, how was I able to solve the Airy equation if I am not qualified?

You have to find the second order differential equation for x(t) and solve it to find x(t). It's the same concept as the algebraic method of finding an equation for x and solving it to get the final answer. I am showing you a way to get to the differential equation. Solving it is up to you.

Your qualifications show themselves in your abilities. This is not as hard a problem as solving the Airy equation.

If you really have or are taking a class in Partial Differential Equations, this really isn't all that hard. So please shut up about the Airy equation and buckle down and take your lumps for making a stupid mistake. It happens, take the blow to your pride like you should, re-learn anything you need to, move on, and don't make the mistake again. This is called "learning" and we've all done it.

Now, where in my three step programme are you getting lost? Take the derivative of the top equation, plug the second equation into it, and use y from the first equation. You'll get a second order equation in x(t). Can you solve that? Write it down, see what you get, and post it if you have any more questions.

-Dan

 

[Steven G]

Why do I think that you are not ready?
Here is why. You think that you are to solve for dy/dt

You are wrong. The question is not asking for $\frac{dx}{dt}$ or $\frac{dy}{dt}$. It asks to solve the system to find $x(t)$ and $y(t)$. If you are confused or mixed this question with something else, please read about Linear Systems. I am telling that I am 100% qualified for this course. Answer this question please, how was I able to solve the Airy equation if I am not qualified?

You mistakenly said that I am wrong. I never said that you should solve for dy/dt or dx/dt. YOU solved for dy/dt--just look at post 1.

 

[mario99]

My method gives you a second order differential equation for x. Solve for x!

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 3y}{6}\right] + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 3y}{6}\right] \right )\right)$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( \frac{2}{6}\dfrac{d^2x}{dt^2} - \frac{4}{6}\dfrac{dx}{dt} - \frac{6}{6}y + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - \frac{2}{6}\dfrac{d^2x}{dt^2} - \frac{4}{6}\dfrac{dx}{dt} - \frac{6}{6}y \right )\right)$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( \frac{2}{6}\dfrac{d^2x}{dt^2} - \frac{4}{6}\dfrac{dx}{dt} - \frac{6}{6}y + \frac{1}{3} \dfrac{dx}{dt} - \frac{2}{18}\dfrac{d^2x}{dt^2} - \frac{4}{18}\dfrac{dx}{dt} - \frac{6}{18}y \right)$

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + \frac{6}{6}\dfrac{d^2x}{dt^2} - \frac{12}{6}\dfrac{dx}{dt} - \frac{18}{6}y + \frac{3}{3} \dfrac{dx}{dt} - \frac{6}{18}\dfrac{d^2x}{dt^2} - \frac{12}{18}\dfrac{dx}{dt} - \frac{18}{18}y$

The differential equation still has y.

Elimination, as in the algebraic case, will not work here.

I am sure 100% elimination must work.

I want you to show me any competent Differential Equations student who does NOT know that the d in dx is not a variable. Otherwise, do you believe that dx/dt = x/t??

Me. I am a super competent differential equation student. This is the first time I know dx is not a multiplication.

You have to find the second order differential equation for x(t) and solve it to find x(t).

I tried but I think that there must be something wrong in your method. My suggestion is to solve in Matrix style. If this is your method of solving linear system, I suppose you don't have an idea about solving the same system with Matrix, don't you? The Matrix is a straightforward method, but the characteristic equation is annoying somehow.

Your qualifications show themselves in your abilities. This is not as hard a problem as solving the Airy equation.

Yeah I beat the Airy equation. Don't worry, that was nothing comparing to the difficult things I have encountered.

If you really have or are taking a class in Partial Differential Equations, this really isn't all that hard. So please shut up about the Airy equation and buckle down and take your lumps for making a stupid mistake.

Why should I shut up about the Airy equation if I have taken a class in partial differential equation? This Airy is very necessary, so Steven or you must not under grade my skills. Any time you or others shrink my high level math, the Airy equation will pop up again to prove the reverse. By the way, I solved it from scratch!

Now, where in my three step programme are you getting lost?

I am not lost. I am just saying I don't understand it fully as I am sure 100% it must be missing something that is letting it not working properly.

Please, look at my steps and let me know if I was missing something.

You mistakenly said that I am wrong. I never said that you should solve for dy/dt or dx/dt. YOU solved for dy/dt--just look at post 1.

No Steven G, at post 1 I solved for x and I tried to solve for y. Not dy/dt.

 

[topsquark]

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \left ( 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 3y}{6}\right] + \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2\left[\frac{\dfrac{d^2x}{dt^2} - 2 \dfrac{dx}{dt} - 3y}{6}\right] \right )\right)$

No. Take it one step at a time.

$\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 \dfrac{dy}{dt}$

Now,
$\dfrac{dy}{dt} = 2x + y$.

So
(Eqn 3) $\dfrac{d^2x}{dt^2} = 2 \dfrac{dx}{dt} + 3 (2x + y)$

From the first equation,
$y = \dfrac{1}{3} \left ( \dfrac{dx}{dt} - 2x \right )$

Sub that into Eqn 3 and simplify.

I am sure 100% elimination must work.

You are wrong and I'm already tired of telling you that.

Me. I am a super competent differential equation student. This is the first time I know dx is not a multiplication.

Then you are not a super competent differential equation student. You are a below average Introductory Calculus student. See here.

I tried but I think that there must be something wrong in your method. My suggestion is to solve in Matrix style. If this is your method of solving linear system, I suppose you don't have an idea about solving the same system with Matrix, don't you? The Matrix is a straightforward method, but the characteristic equation is annoying somehow.

Tell me first: Have you taken Linear Algebra? If not, the matrix method is going to be hopelessly confusing to you. (Don't protest, please just answer the question.)

Yeah I beat the Airy equation. Don't worry, that was nothing comparing to the difficult things I have encountered.

Why should I shut up about the Airy equation if I have taken a class in partial differential equation? This Airy is very necessary, so Steven or you must not under grade my skills. Any time you or others shrink my high level math, the Airy equation will pop up again to prove the reverse. By the way, I solved it from scratch!

You can't even do a simple linear differential vector problem because you are convinced that you know everything. You have no idea how little you actually know. The Airy equation has nothing to do with this problem and your skills are clearly substandard for the material you are trying to work with, as demonstrated by a reasonably elementary four step problem. Your "high level math" is about second semester college Sophomore, and you aren't even that good at it. We are here to teach, you are here to learn (presumably.) Your ego is just getting in the way. You need to go back to Calculus I and do a serious review.

I am a professional educator and you came to this site for help. I am giving you that help: Go back and learn the basics. If you aren't going to listen to that advice, then why are you here? You can find You Tube videos that will teach you how to do this problem and it will save us from wasting your and our time.

Clearly, the new mario is still the old mario. And I'll bet you don't even know why I'm saying that. Pity.

-Dan

 

[khansaheb]

You don't need to tell me the difference between algebraic equation and differential equation.

Then pray enlighten us about the difference between algebraic equation and differential equation.

To the tutors:

Why are we wasting time in this futile exchange.......

 

[mario99]

No. Take it one step at a time.

I will do this in paper. When I finish, I will rewrite it here again.

You are wrong and I'm already tired of telling you that.

I am not wrong and I can show you that if you are interested.

Then you are not a super competent differential equation student. You are a below average Introductory Calculus student. See here.

How could I solve the Airy if I am not super? Missing little things such as dx is not a big deal. Calling me below average is so offensive.

Tell me first: Have you taken Linear Algebra? If not, the matrix method is going to be hopelessly confusing to you. (Don't protest, please just answer the question.)

What do you mean by linear algebra? I am an expert in Matrix. I started studying Matrices when I was in elementary school. Believe it or not I was below 11 years old in that time.

You can't even do a simple linear differential vector problem because you are convinced that you know everything.

I am a professional educator and you came to this site for help.

I didn't say I know everything. I'm not convinced either. From where you get these wrong ideas. Also I don't think a professional educator will talk offensively like what you are doing. Your level of education (teaching) is below average. I give you 4 out of 10. Ten is the best.


Only one professional educator I have encountered in all websites I visited, tkhunny. That man deserves 10 of 10 and a million stars.

 

[stapel]

Okay; that's enough of that. Since you're so much smarter and more advanced that PhD's (and even more so with respect to all the apparently puny brains here), you have no need to waste your time with us. Go "bless" somebody else with your brilliance.

Fly; be free.