Question

So, I need some help with differential equations. I have \[y_1'=-6y_1+7y_2\]\[y_2'=-6y_1+15y_2\], and with the conditions \[\left(\begin{matrix}y_1(0) \\ y_2(0)\end{matrix}\right)=\left(\begin{matrix}5 \\ 0\end{matrix}\right)\] Then I need to find the 2nd coordinatfunction. \[y_2(t)=2e^{-3t}-Oe^{\lambda_2t}\]
This means that I need to find the value for O.

Answer 1

I have in previous questions, found the eigenvectors and values for this system. \[\lambda_1=-3 \]With eigenvector:\[u=\left(\begin{matrix}1 \\ 2\end{matrix}\right) \]\[\lambda_2=12\] With the eigenvector\[v=\left(\begin{matrix}3 \\ 1\end{matrix}\right)\]

Answer 2

In case question's math doesnt load: (it doesnt for me):

So, I need some help with differential equations. I have \[y_1'=-6y_1+7y_2\]\[y_2'=-6y_1+15y_2\], and with the conditions \[\left(\begin{matrix}y_1(0) \\ y_2(0)\end{matrix}\right)=\left(\begin{matrix}5 \\ 0\end{matrix}\right)\] Then I need to find the 2nd coordinatfunction. \[y_2(t)=2e^{-3t}-Oe^{\lambda_2t}\]
This means that I need to find the value for O.

Answer 3

here we can rewrite such system of ODEs, like below:
\[\left( {\begin{array}{*{20}{c}}
{{y_1}'} \\
{{y_2}'}
\end{array}} \right) = \left( {\begin{array}{*{20}{c}}
{ - 6}&7 \\
{ - 6}&{15}
\end{array}} \right) \cdot \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = A\left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right)\]
so, we have to find the eigenvalues of A, first

Answer 4

Is it only me having trouble loading the site sometimes?

Answer 5

please wait a moment, I compute the eigenvalues of A

Answer 6

ohh, wait. It is supposed to be 9, not 7.
In \[y_1'=-6y_1+9y_2\]

Answer 7

Not that it matters that much, just need to understand the general way to do it

Answer 8

ok! Then the eigenvalues are given by the subsequent equation:
\[\Large \begin{gathered}
\det \left( {\begin{array}{*{20}{c}}
{ - 6 - \lambda }&9 \\
{ - 6}&{15 - \lambda }
\end{array}} \right) = 0 \Rightarrow {\lambda ^2} - 9\lambda + 36 = 0 \hfill \\
\hfill \\
{\lambda _1} = - 3,\quad {\lambda _2} = 12 \hfill \\
\end{gathered} \]
and the solution is:
\[\Large \begin{gathered}
{y_1} = {c_1}{e^{ - 3t}} + {c_2}{e^{12t}} \hfill \\
{y_2} = {c_3}{e^{ - 3t}} + {c_4}{e^{12t}} \hfill \\
\end{gathered} \]

Answer 9

oops.. I have made a typo:
\[\Large \det \left( {\begin{array}{*{20}{c}}
{ - 6 - \lambda }&9 \\
{ - 6}&{15 - \lambda }
\end{array}} \right) = 0 \Rightarrow {\lambda ^2} - 9\lambda - 36 = 0\]

Answer 10

Yes, I have found out so much. So now I need to find the values for \[c_4\]

Answer 11

now we have to apply the initial condition, so we can write this:
\[\Large \left\{ \begin{gathered}
5 = {c_1} + {c_2} \hfill \\
0 = {c_3} + {c_4} \hfill \\
\end{gathered} \right.\]

Answer 12

2 equations with 4 variables?

Answer 13

please sorry I have made an error, here are the right equations:

Answer 14

Well, actually I got it now. Because I have given \[c_3\] That would leave \[0=2-c_4<->c_4=2\]

Answer 15

But if I didnt have \[c_3\] would I still be able to solve all the 4 variables with 2 equations?

Answer 16

there are only 2 unknown constants

Answer 17

Are they the same, \[c_1=c_3\] and \[c_2=c_4\]?

Answer 18

also, I got eigenvalue 12 going with vector <1 2> and -3 with <3 1>

Answer 19

Thats the same as I did @phi

Answer 20

here are the solutions:
\[\Large \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}
1 \\
2
\end{array}} \right){e^{ - 3t}} + {c_2}\left( {\begin{array}{*{20}{c}}
3 \\
{21}
\end{array}} \right){e^{12t}}\]

Answer 21

you can think of the answer as
\[ y= c_1 x_1 \exp(\lambda_1 t) + c_2 x_2 \exp(\lambda_2 t)\]
where y, x1 and x2 are all vectors

Answer 22

ah, yea.

Answer 23

Could you help me with a follow up question to this?

Answer 24

sorry another typo:
\[\Large \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}
1 \\
2
\end{array}} \right){e^{ - 3t}} + {c_2}\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right){e^{12t}}\]

Answer 25

Now I have:\[y_1(0)=1\] and\[\lim_{t \rightarrow \infty}y_1(t)=0\] I need to find the missing values: \[y_2(t)=\frac{ 1 }{ 3 }e^{-3t}-(missing)e^{12t}\]

Answer 26

double check. I think it's
\[ \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}
1 \\
2
\end{array}} \right){e^{ 12t}} + {c_2}\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right){e^{-3t}}\]

Answer 27

Yes, that is correct @phi when I insert the eigenvalue -3 I get the vector (3,1)

Answer 28

yes! I got the same result. So the solution, is:
\[\Large \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right){e^{ - 3t}} + {c_2}\left( {\begin{array}{*{20}{c}}
1 \\
2
\end{array}} \right){e^{12t}}\]
now, we have to set \(c_2=0\), if we want \[{y_1} \to 0\]

Answer 29

when t---> +infinity

Answer 30

I am not sure I follow, \[c_2\] in which equation?

Answer 31

sorry, I meant in my last equation

Answer 32

\[\Large \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right){e^{ - 3t}} + {c_2}\left( {\begin{array}{*{20}{c}}
1 \\
2
\end{array}} \right){e^{12t}}\]

Answer 33

since we have:
\[\mathop {\lim }\limits_{x \to + \infty } {e^{12t}} = + \infty \]

Answer 34

We have \[\lim_{t \rightarrow \infty}y_1(t)=0\]

Answer 35

since
\[\Large {e^{ - 3t}},\;\;{e^{12t}}\]
are two linearly independent functions, we have to set \(c_2=0\), otherwise, we wiil not get such condition you wrote

Answer 36

will*

Answer 37

Yes, how do I translate that into \[y_2(t)\]?

Answer 38

we get this:
\[\Large \left( {\begin{array}{*{20}{c}}
{{y_1}} \\
{{y_2}}
\end{array}} \right) = {c_1}\left( {\begin{array}{*{20}{c}}
3 \\
1
\end{array}} \right){e^{ - 3t}} = \left( {\begin{array}{*{20}{c}}
{3{c_1}{e^{ - 3t}}} \\
{{c_1}{e^{ - 3t}}}
\end{array}} \right)\]
now, I can write:
\[\Large 3{c_1} = 1 \Rightarrow {c_1} = \frac{1}{3}\]

Answer 39

and, of course:
\[\Large {\text{missing value}} = {c_2} = 0\]

Answer 40

ah, yea. Thank you SO much :)

Answer 41

:)