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- Apr 13, 2013

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

How can I do this??

- Thread starter evinda
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[tex] \lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6} [/tex]

How can I do this??

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- Mar 5, 2012

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Hi evinda!

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

How can I do this??

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 $x \to \infty$?

- Feb 29, 2012

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Have you found the solutions to both differential equations?

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

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

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

Last edited:

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I think that u is not equal to [tex] y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) [/tex] and v not to [tex] y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) [/tex] .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?I found that the solution of the differential equation [tex]y''-4y'+29y=0[/tex] is [tex] y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) [/tex] and that the solution of the differential equation [tex] y''+4y'+13y=0[/tex] is [tex] y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) [/tex] .

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

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

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

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

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

To find u and v you have to use the boundary condition (0,0).I think that u is not equal to [tex] y(x)=c_{1}e^{2x}sin(5x)+c_{2}e^{2x}cos(5x) [/tex] and v not to [tex] y(x)=d_{1}e^{-2x}sin(3x)+d_{2}e^{-2x}cos(3x) [/tex] .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?

That is, u(0)=0 and v(0)=0.

After that you can determine the limit.

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- Apr 13, 2013

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I used it and found that [tex] u(x)=c_{1}e^{2x}sin(5x) [/tex] and [tex] v(x)=d_{1}e^{-2x}sin(3x) [/tex] .And now?how can I continue?Good!

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.

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Very good.I used it and found that [tex] u(x)=c_{1}e^{2x}sin(5x) [/tex] and [tex] v(x)=d_{1}e^{-2x}sin(3x) [/tex].

Substitute in \(\displaystyle \lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}\)?And now?how can I continue?

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It is equal to [tex] \lim_{x\rightarrow \infty}\frac{d_{1}^{4}e^{(-2x)^{4}}sin^{4}(3x)}{c_{1}e^{2x}sin(5x)} [/tex] ,right?But can this limit[tex] \rightarrow \frac{5}{6} [/tex] ?Very good.

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

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Correct!It is equal to [tex] \lim_{x\rightarrow \infty}\frac{d_{1}^{4}e^{(-2x)^{4}}sin^{4}(3x)}{c_{1}e^{2x}sin(5x)} [/tex] ,right?

Neh. It can't.But can this limit[tex] \rightarrow \frac{5}{6} [/tex] ?

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Correct!

Neh. It can't.

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

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

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

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Yep. That's what I think.The exercise asks me to find u,v so that:

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

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

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I am looking again at the exercise and realized that I didn't write right the differential equations. $u$ 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?Yep. That's what I think.

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You solved it before... can you solve it again?I am looking again at the exercise and realized that I didn't write right the differential equations. $u$ 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?

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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 $v(x)=d_{1}e^{x}+d_{2}e^{-5x}$.You solved it before... can you solve it again?

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?? But how can I find $u,v$ so that $ \lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?

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Substitute?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?? But how can I find $u,v$ so that $ \lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?

You may limit yourself to the 2 highest powers of $e^x$ in $v^4(x)$.

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- #18

- Apr 13, 2013

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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?Substitute?

You may limit yourself to the 2 highest powers of $e^x$ in $v^4(x)$.

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}}

$ ?

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That looks about right.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}}

$ ?

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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??That looks about right.

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The same principle applies. The other powers go to zero so fast that they have no impact on the limit.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??

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.

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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?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.

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So let's start with 1 solution.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?

Suppose we pick $d_1=0$, does the corresponding solution for u and v satisfy all conditions?

What if $d_1=1$?

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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 $0$ ?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 $d_1=1$?

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Well... you should also check that the differential equations and the boundary conditions are satisfied...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 $0$ ?