# [SOLVED]3.2.5.3 General Solution if system

#### karush

##### Well-known member
$\tiny{307w.3.2.5.3}$
source
Find the general solution of each of the linear system
\begin{align*}
x' & = -3 x + 4y\\
y' & = 3x - 2y
\end{align*}
$A=\begin{pmatrix}-3&4\\ 3&-2\end{pmatrix} =\left[\begin{array}{rr}- \lambda - 3 & 4\\3 & - \lambda - 2\end{array}\right] =\lambda^{2} + 5 \lambda - 6 = 0 \quad \lambda_1=-5\quad \lambda_2=6$
\textit{ eigenvector:}$\left[ \begin{array}{r}1\\1\end{array}\right]$
and eigenvector:
$\left[\begin{array}{r}- \frac{4}{3}\\1\end{array}\right]$
so
$Au=\begin{pmatrix}-3&4\\ 3&-2\end{pmatrix} \left(\begin{array}{r}1\\1\end{array}\right) =\left(\begin{array}{r}1\\1\end{array}\right) =\lambda_1\left(\begin{array}{r}1\\1\end{array}\right) =e^{-5t}\left[ \begin{array}{r}1\\1\end{array}\right]$

so far.... but need other GE
typos probably

tried doing it w/o matrix but

#### Cbarker1

##### Active member
$\tiny{307w.3.2.5.3}$
source
Find the general solution of each of the linear system
\begin{align*}
x' & = -3 x + 4y\\
y' & = 3x - 2y
\end{align*}
$A=\begin{pmatrix}-3&4\\ 3&-2\end{pmatrix} =\left[\begin{array}{rr}- \lambda - 3 & 4\\3 & - \lambda - 2\end{array}\right] =\lambda^{2} + 5 \lambda - 6 = 0 \quad \lambda_1=-5\quad \lambda_2=6$
\textit{ eigenvector:}$\left[ \begin{array}{r}1\\1\end{array}\right]$
and eigenvector:
$\left[\begin{array}{r}- \frac{4}{3}\\1\end{array}\right]$
so
$Au=\begin{pmatrix}-3&4\\ 3&-2\end{pmatrix} \left(\begin{array}{r}1\\1\end{array}\right) =\left(\begin{array}{r}1\\1\end{array}\right) =\lambda_1\left(\begin{array}{r}1\\1\end{array}\right) =e^{-5t}\left[ \begin{array}{r}1\\1\end{array}\right]$

so far.... but need other GE
typos probably

tried doing it w/o matrix but
What is the vector $b(t)=\begin{pmatrix} x(t)\\ y(t) \end{pmatrix}$ equal to? So what is the $b'(t)$? What is the other eigenvector and eigenvalue? How does the other eigenvector and eigenvalue relate to your first part of the solution to this system of ODEs? What is the superposition principle in second order ODE and how does that principle relates to the systems of ODE?

I have a hard time understanding your notation because I think you are using the equal sign as an operator, which is not the correct usage. I like that you tell me what the matrix $A$ is. What is the vector $u$ states for?
But if you are using eigenvalue algorithm, you need to say $\det(A-\lambda I_2)=$.. then compute the roots of the characteristic polynomial of the matrix $A$ on a different line.

#### karush

##### Well-known member
this is the example I am trying to follow

#### Cbarker1

##### Active member
What
this is the example I am trying to follow

View attachment 11104
What do the third paragraph read, "similiarly,..."? Have you try to plugin in the second eigenvector to see if it is associated to the second eigenvalue? I.e. Repeated this line
$Au=\begin{pmatrix}-3&4\\ 3&-2\end{pmatrix} \left(\begin{array}{r}1\\1\end{array}\right) =\left(\begin{array}{r}1\\1\end{array}\right) =\lambda_1\left(\begin{array}{r}1\\1\end{array}\right) =e^{-5t}\left[ \begin{array}{r}1\\1\end{array}\right]$ with the second eigenvector. To see if the eigenvalue is the same?

#### karush

##### Well-known member
probably not those are steps i don't understand

#### Cbarker1

##### Active member
Can you do matrix multiplication?
$$\begin{pmatrix}-3&4\\ 3&-2\end{pmatrix} \left(\begin{array}{r}1\\1\end{array}\right) =\left(\begin{array}{r}1\\1\end{array}\right)$$

#### Cbarker1

##### Active member
I think that your eigenvalues are incorrect. Let's restart from the beginning in order for you to understand the idea is. Let's find the eigenvalue of your ODEs. What is the $det(\lambda I_2-A)=$, where $A$ is your matrix that is defined and $I_2$ is the identity matrix?

#### karush

##### Well-known member
I think that your eigenvalues are incorrect. Let's restart from the beginning in order for you to understand the idea is. Let's find the eigenvalue of your ODEs. What is the $det(\lambda I_2-A)=$, where $A$ is your matrix that is defined and $I_2$ is the identity matrix?
here is where I got the eigenvalues
eMh

#### Cbarker1

##### Active member
here is where I got the eigenvalues
eMh
Ok. Let's use that calculator as way to check it. Are you familiar of determinants and quadratic formula?

#### Country Boy

##### Well-known member
MHB Math Helper
$\lambda$ is an "eigenvalue" of linear transformation, A, if and only if there exist a non-zero vector, v, such that $Av= \lambda v$.

So you want to solve $\begin{bmatrix}-3 & 4 \\ 3 & -2\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= \lambda\begin{bmatrix} x \\ y \end{bmatrix}$. That is the same as $\begin{bmatrix}-3 & 4 \\ 3 & -2 \end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}- \lambda\begin{bmatrix} x \\ y\end{bmatrix}= \begin{bmatrix}-3-\lambda & 4 \\ 3 & -2-\lambda\end{bmatrix}\begin{bmatrix}x \\ y \end{bmatrix}= 0$.

An obvious solution to that is x= 0, y= 0 but we are looking for non-zero (non-trivial) solutions. As a general rule, if A is invertible, which for a matrix is equivalent to saying its determinant is not 0, Ax= b has the unique solution $x= A^{-1}b$ and if b= 0, x= 0 would be the only solution.

So to have a non-zero solution, we must have $\left|\begin{array}{cc}-3- \lambda & 4 \\ 3 & -2- \lambda \end{array}\right|= (-3-\lambda)(-2-\lambda)- 4(3)= \lambda^2+ 5\lambda+ 6- 12= \lambda^2+ 5\lambda- 6= (\lambda+ 6)(\lambda- 1)= 0$.

The eigenvalues are $\lambda= -6$ and $\lambda= 1$.

Now, we look for the corresponding eigenvectors. Again we go back to the basic definition. $v= \begin{bmatrix} x \\ y \end{bmatrix}$ is an eigenvector corresponding to the eigenvalue $\lambda= 1$ if and only if $\begin{bmatrix}-3 & 4 \\ 3 & -2 \end{bmatrix}$$\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-3x+ 4y \\ 3x- 2y \end{bmatrix}= 1 \begin{bmatrix}x \\ y \end{bmatrix}. That is equivalent to the two equations -3x+ 4y= x, or -4x+ 4y= 0 and 3x- 2y= y or 3x- 3y= 0, Those both reduce to y= x so any vector \begin{bmatrix} x \\ x \end{bmatrix}= x\begin{bmatrix} 1 \\ 1\end{bmatrix}. Yes, there are an infinite number of eigenvectors corresponding to eigenvalue 1: the subspace spanned by \begin{bmatrix} 1 \\ 1 \end{bmatrix}, Now, for eigenvectors corresponding to eigenvalue -6. Same basic idea- we want \begin{bmatrix} x \\ y \end{bmatrix} such that \begin{bmatrix}-3 & 4 \\ 3 & -2 \end{bmatrix}$$\begin{bmatrix}x \\ y \end{bmatrix}= \begin{bmatrix}-3x+ 4y \\ 3x- 2y \end{bmatrix}= -6 \begin{bmatrix}x \\ y \end{bmatrix}$ which is equivalent to the two equations $-3x+ 4y= -6x$ or $3x+ 4y= 0$ and $3x-2y= -6y$ or $3x+ 4y= 0$. Again, there are infinitely many non-zero solution to that. Taking x= 4 and y= -3 satisfy that- as well as any multiple of $\begin{bmatrix} 4 \\ -3 \end{bmatrix}$. All vectors in the subspace spanned by $\begin{bmatrix}4 \\ -3 \end{bmatrix}$ are eigenvectors corresponding to the eigenvalue -6.

#### karush

##### Well-known member
wow, that was a lot of explanation but was a very great help.
I was having trouble with setting up the matrix's
i still don't see how this makes the derivative disappear

#### Cbarker1

##### Active member
It is like the second order ODE...when you solve them, you make a guess of what the solution will look like.

For instance, $y''+5y'+6y=0$ the only function that has this property is $e^{rx}$.

#### karush

##### Well-known member
well that's good to know...

Beach House one of last projects I worked on before I retired

w

#### Cbarker1

##### Active member
well that's good to know...

Beach House one of last projects I worked on before I retired

wView attachment 11105
Since system of ode will have n equations, the system will be translated to nth order ode (Depends on the system if the ode is linear or not; or have constant coefficients or not).

#### karush

##### Well-known member
ok I started another thread with a system of 2 eq with IV's