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- #26

- Mar 5, 2012

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- Thread starter evinda
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- Admin
- #26

- Mar 5, 2012

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- Thread starter
- #27

- Apr 13, 2013

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- Thread starter
- #28

- Apr 13, 2013

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

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- #29

- Mar 5, 2012

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

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

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?

For $d_1 = 1$, we get $d_2=-1$, and therefore $c_1 = 6/5$ and $c_2 = -6/5$.From the boundary conditions don't we get that $c_{1}=\frac{2d_{1}-4d_{2}}{5} \text{ and } c_{2}=\frac{3d_{1}+9d_{2}}{5}$ ? And how can I check if the differential equations are satisfied?Aren't they satisfied for any $d_{1},d_{2},c_{1},c_{2}$ ?

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!!!

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- #30

- Apr 13, 2013

- 3,828

Nice!!Thank you very much!!!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!!!