MHB Find u,v of the differential equation

[evinda]

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??

 

[I like Serena]

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! (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 $x \to \infty$?

 

[Fantini]

Re: Find u,v!

Have you found the solutions to both differential equations? :)

 

[evinda]

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

 

[evinda]

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?

 

[I like Serena]

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! (Smile)

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]

Re: Find u,v!

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?

 

[I like Serena]

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

 

[evinda]

Re: Find u,v!

Very good.

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} ?

 

[I like Serena]

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]

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?

 

[I like Serena]

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.

 

[evinda]

Re: Find u,v!

Yep. That's what I think.

Ok...Thank you :)

 

[evinda]

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?

 

[I like Serena]

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? (Thinking)

 

[evinda]

Re: Find u,v!

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

 

[I like Serena]

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]

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}}
$ ?

 

[I like Serena]

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}}
$ ?

That looks about right. :)

 

[evinda]

Re: Find u,v!

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)

 

[I like Serena]

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?? (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.

 

[evinda]

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?

 

[I like Serena]

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?

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 $d_1=1$?

 

[evinda]

Re: Find u,v!

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 $0$ ?

 

[I like Serena]

Re: Find u,v!

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...

 

[I like Serena]

Oh, and:

View attachment 1894

(Worried) (Bandit) (Evilgrin)

 

[evinda]

Oh, and:

View attachment 1894

(Worried) (Bandit) (Evilgrin)

(Giggle)(Bigsmile)(Wasntme)

 

[evinda]

Re: Find u,v!

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 $d_{1},d_{2},c_{1},c_{2}$ ?

 

[I like Serena]

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??

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}$ ?

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?

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)

 

[evinda]

Re: Find u,v!

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! :)