# Find u,v!

#### evinda

##### Well-known member
MHB Site Helper
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??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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!

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 $x \to \infty$?

#### Fantini

MHB Math Helper
Re: Find u,v!!!

Have you found the solutions to both differential equations?

#### evinda

##### Well-known member
MHB Site Helper
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

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

Last edited:

#### evinda

##### Well-known member
MHB Site Helper
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

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

#### Klaas van Aarsen

##### MHB Seeker
Staff member
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

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

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.

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

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.
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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.

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

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

Very good.

Substitute in $$\displaystyle \lim_{x\rightarrow \infty}\frac{v^{4}(x)}{u(x)}$$?
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$$\rightarrow \frac{5}{6}$$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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$$\rightarrow \frac{5}{6}$$ ?
Neh. It can't.

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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.

MHB Site Helper

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

Yep. That's what I think.
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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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?
You solved it before... can you solve it again?

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

You solved it before... can you solve it again?
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 $\lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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 $\lim_{x\to\infty}\frac{v^{4}(x)}{u(x)}=\frac{5}{6}$ ?
Substitute?
You may limit yourself to the 2 highest powers of $e^x$ in $v^4(x)$.

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

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:$\lim_{x \to \infty} \frac{d_{1}^{4}e^{4x}}{\frac{2d_{1}-4d_{2}}{5}e^{4x}}=\frac{d_{1}^{4}}{\frac{2d_{1}-4d_{2}}{5}}$
?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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:$\displaystyle \lim_{x \to \infty} \frac{d_{1}^{4}e^{4x}}{\frac{2d_{1}-4d_{2}}{5}e^{4x}}=\frac{d_{1}^{4}}{\frac{2d_{1}-4d_{2}}{5}}$ ?

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

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??

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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??
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.

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

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?

#### Klaas van Aarsen

##### MHB Seeker
Staff member
Re: Find u,v!!!

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 $d_1=1$?

#### evinda

##### Well-known member
MHB Site Helper
Re: Find u,v!!!

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 $0$ ?
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 $0$ ?