Find u,v of the differential equation

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Discussion Overview

The discussion revolves around finding the functions u and v that satisfy specific differential equations and intersect at the point (0,0). The participants explore the implications of a limit condition involving these functions as x approaches infinity, while also addressing potential misprints in the original problem statement.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants inquire about solving the differential equations y''-4y'+29y=0 and y''+4y'+13y=0, and whether the limit condition should be evaluated as x approaches 0 instead of infinity.
  • One participant provides the solutions to the differential equations and expresses uncertainty about the limit condition given in the exercise.
  • Another participant suggests that the limit condition cannot be satisfied with the initially proposed solutions for u and v.
  • Later, a participant realizes that the differential equations were incorrectly stated and provides the correct equations, prompting further discussion about the limit's existence.
  • Participants discuss the significance of the highest powers of e in the limit calculation and whether it is appropriate to consider only the highest powers in the context of exponential functions.

Areas of Agreement / Disagreement

There is no consensus on whether the original limit condition can be satisfied with the initially proposed solutions. The discussion evolves as participants correct the differential equations, leading to further exploration of the limit condition.

Contextual Notes

Participants express uncertainty regarding the limit condition and the implications of the boundary conditions. There are also discussions about the relevance of the highest powers in the limit calculation, indicating a nuanced understanding of the mathematical principles involved.

evinda
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If the graph of the solution u of the differential equation y''-4y'+29y=0 intersects the graph of the solution v of the differential equation y''+4y'+13y=0 at the point (0,0),find u,v so that:

\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}

How can I do this??
 
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Re: Find u,v!

evinda said:
If the graph of the solution u of the differential equation y''-4y'+29y=0 intersects the graph of the solution v of the differential equation y''+4y'+13y=0 at the point (0,0),find u,v so that:

\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}

How can I do this??

Hi evinda! (Smirk)

Did you try and solve those DE's with the given boundary condition?

Btw, can it be that the limit should let $x \to 0$ instead of $x \to \infty$?
 
Re: Find u,v!

Have you found the solutions to both differential equations? :)
 
Re: Find u,v!

I found that the solution of the differential equation y''-4y'+29y=0 is y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) and that the solution of the differential equation y''+4y'+13y=0 is y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) .

At the exercise I am looking at,it says that x\rightarrow \infty,but if you think that it might be wrong,it could be a misprint :confused:

I tried to find the limit but I don't know what to do to find u and v :confused:
 
Last edited:
Re: Find u,v!

evinda said:
I found that the solution of the differential equation y''-4y'+29y=0 is y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) and that the solution of the differential equation y''+4y'+13y=0 is y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) .

At the exercise I am looking at,it says that x\rightarrow \infty,but if you think that it might be wrong,it could be a misprint :confused:

I tried to find the limit but I don't know what to do to find u and v :confused:

I think that u is not equal to y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) and v not to y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) .Because if it was like that,we wouldn't use the limit that is given from the exercise..Or am I wrong??But,how else could I find u and v?
 
Re: Find u,v!

evinda said:
I found that the solution of the differential equation y''-4y'+29y=0 is y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) and that the solution of the differential equation y''+4y'+13y=0 is y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) .

At the exercise I am looking at,it says that x\rightarrow \infty,but if you think that it might be wrong,it could be a misprint :confused:

I tried to find the limit but I don't know what to do to find u and v :confused:

Good! (Smile)
evinda said:
I think that u is not equal to y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) and v not to y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) .Because if it was like that,we wouldn't use the limit that is given from the exercise..Or am I wrong??But,how else could I find u and v?

To find u and v you have to use the boundary condition (0,0).
That is, u(0)=0 and v(0)=0.

After that you can determine the limit.
 
Re: Find u,v!

I like Serena said:
Good! (Smile)

To find u and v you have to use the boundary condition (0,0).
That is, u(0)=0 and v(0)=0.

After that you can determine the limit.

I used it and found that u(x)=c_{1}e^{2x}sin(5x) and v(x)=d_{1}e^{-2x}sin(3x) .And now?how can I continue?
 
Re: Find u,v!

evinda said:
I used it and found that u(x)=c_{1}e^{2x}sin(5x) and v(x)=d_{1}e^{-2x}sin(3x).

Very good. :cool:
And now?how can I continue?

Substitute in $$\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}$$?
 
Re: Find u,v!

I like Serena said:
Very good. :cool:

Substitute in $$\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}$$?

It is equal to \lim_{x\rightarrow \infty}\frac{d_{1}^{4}e^{(-2x)^{4}}sin^{4}(3x)}{c_{1}e^{2x}sin(5x)} ,right?But can this limit\rightarrow \frac{5}{6} ?:confused:
 
  • #10
Re: Find u,v!

evinda said:
It is equal to \lim_{x\rightarrow \infty}\frac{d_{1}^{4}e^{(-2x)^{4}}sin^{4}(3x)}{c_{1}e^{2x}sin(5x)} ,right?

Correct!
But can this limit\rightarrow \frac{5}{6} ?:confused:

Neh. It can't.
 
  • #11
Re: Find u,v!

I like Serena said:
Correct!

Neh. It can't.
The exercise asks me to find u,v so that:

\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}

So,is the right answer that we can't find u,v with this condition?
 
  • #12
Re: Find u,v!

evinda said:
The exercise asks me to find u,v so that:

\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}

So,is the right answer that we can't find u,v with this condition?

Yep. That's what I think.
 
  • #13
Re: Find u,v!

I like Serena said:
Yep. That's what I think.

Ok...Thank you :)
 
  • #14
Re: Find u,v!

I like Serena said:
Yep. That's what I think.

I am looking again at the exercise and realized that I didn't write right the differential equations. :o $u$ is the solution of the differential equation $y''-3y'-4y=0$,and $v$ the solution of this: $y''+4y'-5y=0$,u and v intersect at the point (0,0) ,where they have also the same deviation.Could you tell me,if the limit exists now? :o
 
  • #15
Re: Find u,v!

evinda said:
I am looking again at the exercise and realized that I didn't write right the differential equations. :o $u$ is the solution of the differential equation $y''-3y'-4y=0$,and $v$ the solution of this: $y''+4y'-5y=0$,u and v intersect at the point (0,0) ,where they have also the same deviation.Could you tell me,if the limit exists now? :o

You solved it before... can you solve it again? (Thinking)
 
  • #16
Re: Find u,v!

I like Serena said:
You solved it before... can you solve it again? (Thinking)

I found that the solution of the differential equation $y''-3y'-4y=0$ is $u(x)=c_{1}e^{4x}+c_{2}e^{-x}$ and that of this differential equation $y''+4y'-5y=0$ it is $v(x)=d_{1}e^{x}+d_{2}e^{-5x}$.
Then from these conditions : $u(0)=v(0) \text { and } u'(0)=v'(0) $ I found that $c_{1}=\frac{2d_{1}-4d_{2}}{5} \text{ and } c_{2}=\frac{3d_{1}+9d_{2}}{5}$ .Is this right??:confused: But how can I find $u,v$ so that $ \lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?
 
  • #17
Re: Find u,v!

evinda said:
I found that the solution of the differential equation $y''-3y'-4y=0$ is $u(x)=c_{1}e^{4x}+c_{2}e^{-x}$ and that of this differential equation $y''+4y'-5y=0$ it is $v(x)=d_{1}e^{x}+d_{2}e^{-5x}$.
Then from these conditions : $u(0)=v(0) \text { and } u'(0)=v'(0) $ I found that $c_{1}=\frac{2d_{1}-4d_{2}}{5} \text{ and } c_{2}=\frac{3d_{1}+9d_{2}}{5}$ .Is this right??:confused: But how can I find $u,v$ so that $ \lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?

Substitute?
You may limit yourself to the 2 highest powers of $e^x$ in $v^4(x)$.
 
  • #18
Re: Find u,v!

I like Serena said:
Substitute?
You may limit yourself to the 2 highest powers of $e^x$ in $v^4(x)$.

The limit is now $\lim_{x \to \infty} \frac{(d_{1}e^{x}+d_{2}e^{-5x})^{4}}{(\frac{2d_{1}-4d_{2}}{5})e^{4x}+(\frac{3d_{1}+9d_{2}}{5})e^{-x}}$,right?
So,you mean that it is equal to thi one:$\lim_{x \to \infty} \frac{d_{1}^{4}e^{4x}}{\frac{2d_{1}-4d_{2}}{5}e^{4x}}=\frac{d_{1}^{4}}{\frac{2d_{1}-4d_{2}}{5}}
$ ?
 
  • #19
Re: Find u,v!

evinda said:
The limit is now $\displaystyle \lim_{x \to \infty} \frac{(d_{1}e^{x}+d_{2}e^{-5x})^{4}}{(\frac{2d_{1}-4d_{2}}{5})e^{4x}+(\frac{3d_{1}+9d_{2}}{5})e^{-x}}$,right?
So,you mean that it is equal to thi one:$\displaystyle \lim_{x \to \infty} \frac{d_{1}^{4}e^{4x}}{\frac{2d_{1}-4d_{2}}{5}e^{4x}}=\frac{d_{1}^{4}}{\frac{2d_{1}-4d_{2}}{5}}
$ ?

That looks about right. :)
 
  • #20
Re: Find u,v!

I like Serena said:
That looks about right. :)

But...why do we take the two highest powers?? Don't we take the highest powers,only when we have a fraction with polynomials?? Or am I wrong?? (Thinking)
 
  • #21
Re: Find u,v!

evinda said:
But...why do we take the two highest powers?? Don't we take the highest powers,only when we have a fraction with polynomials?? Or am I wrong?? (Thinking)

The same principle applies. The other powers go to zero so fast that they have no impact on the limit.
Actually, it suffices to only look at the highest power.
Divide both numerator and denominator by this highest power and the reason should become clear.
 
  • #22
Re: Find u,v!

I like Serena said:
The same principle applies. The other powers go to zero so fast that they have no impact on the limit.
Actually, it suffices to only look at the highest power.
Divide both numerator and denominator by this highest power and the reason should become clear.

I understand.. :) I found that $-2d_{2}=3d_{1}^{4}-d_{1}$ .Is this right?So,I can't find a unique solution for $d_{1},d_{2},c_{1},c_{2}$,or is it possible?
 
  • #23
Re: Find u,v!

evinda said:
I understand.. :) I found that $-2d_{2}=3d_{1}^{4}-d_{1}$ .Is this right?So,I can't find a unique solution for $d_{1},d_{2},c_{1},c_{2}$,or is it possible?

So let's start with 1 solution.
Suppose we pick $d_1=0$, does the corresponding solution for u and v satisfy all conditions?
What if $d_1=1$?
 
  • #24
Re: Find u,v!

I like Serena said:
So let's start with 1 solution.
Suppose we pick $d_1=0$, does the corresponding solution for u and v satisfy all conditions?
What if $d_1=1$?

If we have $d_{1}=0$,then $d_{2}=0$,so the denominator will also be equal to $0$ ,something that is not allowed! For $d_{1}=1$ we find that $d_{2}=-1$,and then the limit will be equal to $\frac{5}{6}$..So,can we take any number for $d_{1}$, apart from $0$ ?
 
  • #25
Re: Find u,v!

evinda said:
If we have $d_{1}=0$,then $d_{2}=0$,so the denominator will also be equal to $0$ ,something that is not allowed! For $d_{1}=1$ we find that $d_{2}=-1$,and then the limit will be equal to $\frac{5}{6}$..So,can we take any number for $d_{1}$, apart from $0$ ?

Well... you should also check that the differential equations and the boundary conditions are satisfied...
 
  • #26

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  • #27
I like Serena said:
Oh, and:

View attachment 1894

(Worried) (Bandit) (Evilgrin)

(Giggle)(Bigsmile)(Wasntme)
 
  • #28
Re: Find u,v!

I like Serena said:
Well... you should also check that the differential equations and the boundary conditions are satisfied...

From the boundary conditions don't we get that $c_{1}=\frac{2d_{1}-4d_{2}}{5} \text{ and } c_{2}=\frac{3d_{1}+9d_{2}}{5}$ ? (Thinking) And how can I check if the differential equations are satisfied?Aren't they satisfied for any $d_{1},d_{2},c_{1},c_{2}$ ?
 
  • #29
Re: Find u,v!

evinda said:
If the graph of the solution u of the differential equation y''-4y'+29y=0 intersects the graph of the solution v of the differential equation y''+4y'+13y=0 at the point (0,0),find u,v so that:

\lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}

How can I do this??

evinda said:
I found that the solution of the differential equation $y''-3y'-4y=0$ is $u(x)=c_{1}e^{4x}+c_{2}e^{-x}$ and that of this differential equation $y''+4y'-5y=0$ it is $v(x)=d_{1}e^{x}+d_{2}e^{-5x}$.
Then from these conditions : $u(0)=v(0) \text { and } u'(0)=v'(0) $ I found that $c_{1}=\frac{2d_{1}-4d_{2}}{5} \text{ and } c_{2}=\frac{3d_{1}+9d_{2}}{5}$ .Is this right??:confused: But how can I find $u,v$ so that $ \lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?

evinda said:
I understand.. :) I found that $-2d_{2}=3d_{1}^{4}-d_{1}$ .Is this right?So,I can't find a unique solution for $d_{1},d_{2},c_{1},c_{2}$,or is it possible?

evinda said:
From the boundary conditions don't we get that $c_{1}=\frac{2d_{1}-4d_{2}}{5} \text{ and } c_{2}=\frac{3d_{1}+9d_{2}}{5}$ ? (Thinking) And how can I check if the differential equations are satisfied?Aren't they satisfied for any $d_{1},d_{2},c_{1},c_{2}$ ?

For $d_1 = 1$, we get $d_2=-1$, and therefore $c_1 = 6/5$ and $c_2 = -6/5$.

So the solutions are:
$$u(x)=\frac 6 5 e^{4x} - \frac 6 5 e^{-x}$$
$$v(x)=e^{x}-e^{-5x}$$

Substituting in $u''-3u'-4u=0$ gives:
$$\left(\frac 6 5 4^2 - 3 \frac 6 5 4 - 4 \frac 6 5\right)e^{4x} - \left(\frac 6 5 + 3\frac 6 5 - 4 \frac 6 5\right)e^{-x} = 0$$
Yay! We verified that this u(x) is actually a solution!
We have confirmation that there is no calculation mistake and that are no weird conditions either.

Then $v(x)$ will probably also be a solution of $v''+4v'-5v=0$.

Furthermore, $u(0) = 0$, which is the same as $v(0)=0$. Also good.
And $u'(0)=\frac 6 5 \cdot 4 + \frac 6 5 = 6$, while $v'(0) = 1 + 5 = 6$.
Yay! More confirmation.So yes, we've got a solution for any $d_1$ with the condition that $2d_1 - 4d_2 \ne 0$ where $−2d_2=3d_1^4−d_1$. Solving that leads indeed to $d_1 \ne 0$.
Yay! It's all good!
Good that you didn't make any calculation mistakes! (Cool)
 
  • #30
Re: Find u,v!

I like Serena said:
For $d_1 = 1$, we get $d_2=-1$, and therefore $c_1 = 6/5$ and $c_2 = -6/5$.

So the solutions are:
$$u(x)=\frac 6 5 e^{4x} - \frac 6 5 e^{-x}$$
$$v(x)=e^{x}-e^{-5x}$$

Substituting in $u''-3u'-4u=0$ gives:
$$\left(\frac 6 5 4^2 - 3 \frac 6 5 4 - 4 \frac 6 5\right)e^{4x} - \left(\frac 6 5 + 3\frac 6 5 - 4 \frac 6 5\right)e^{-x} = 0$$
Yay! We verified that this u(x) is actually a solution!
We have confirmation that there is no calculation mistake and that are no weird conditions either.

Then $v(x)$ will probably also be a solution of $v''+4v'-5v=0$.

Furthermore, $u(0) = 0$, which is the same as $v(0)=0$. Also good.
And $u'(0)=\frac 6 5 \cdot 4 + \frac 6 5 = 6$, while $v'(0) = 1 + 5 = 6$.
Yay! More confirmation.So yes, we've got a solution for any $d_1$ with the condition that $2d_1 - 4d_2 \ne 0$ where $−2d_2=3d_1^4−d_1$. Solving that leads indeed to $d_1 \ne 0$.
Yay! It's all good!
Good that you didn't make any calculation mistakes! (Cool)

Nice!Thank you very much! :)
 

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