[SOLVED]307.27.1

karush

Well-known member
$\tiny{27.1}$
Find a general solution to the system of differential equations
$\begin{array}{llrr}\displaystyle \textit{given} &y'_1=\ \ y_1+2y_2\\ &y'_2=3y_1+2y_2\\ \textit{solving } &A=\begin{pmatrix}1 &2\\3 &2\end{pmatrix}\\ \textit{eigensystem}. &\begin{pmatrix}1-\lambda &2\\3 &2-\lambda\end{pmatrix} =\lambda^2-3\lambda -4 = (\lambda-4)(\lambda+1) = 0 \\ &\lambda = 4,-1 \end{array}$

so far,,, not sure what is next!

Country Boy

Well-known member
MHB Math Helper
Okay, why did you find the eigenvalues of that matrix? What was your purpose?

For something as simple as this, I wouldn't use "matrices" at all. That is too "sophisticated" for me!

The two equations are
$y_1'= y_1+ 2y_2$ and
$y_2'= 3y_1+ 2y_2$.

Differentiate the first equation again:
$y_1''= y_1'+ 2y_2'$.
Substitue for $y_2'$ from the second equation:
$y_1''= y_1'+ 2(3y_1+ 2y_2)= 6y_1+ 4y_2$.
From the first equation $2y_2= y_1'- y_1$ so that is
$y_1''= y_1'+ 6y_1+ 2y_1'- 2y_1= 3y_1'- 4y_1$
$y_1''- 3y_1'- 4y_1= 0$.

That has "characteristic equation $r^2- 3r- 4= (r- 4)(r+ 1)= 0$ with "characteristic values" r= -1 and r= 4. Notice that those are the same as the "eigenvalue equation" and the "eigenvalues". The general solution is $y_1(x)= Ae^{-x}+ Be^{4x}$. You can get $y_2$ from the equation $2y_2= y_1'- y_1$:
$2y_2= -Ae^{-x}+ 4Be^{4x}- Ae^{-x}- Be^{4x}= -2Ae^{-x}+ 3Be^{4x}$.

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Country Boy

Well-known member
MHB Math Helper
Now, as to the solution using a matrix equation.

The problem is to solve $\frac{d\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dx}= \begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}$.

You have determined (correctly) that the eigenvalues of the coefficient matrix are 4 and -1. That means that the matrix can be "diagonalized". That is, there exists a matrix, P, such that $P\begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}P^{-1}= \begin{pmatrix}4 & 0\\0 & -1\end{pmatrix}$.

And, of course, then. that $\begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}= P^{-1}\begin{pmatrix}4 & 0 \\ 0 & -1 \end{pmatrix}P$.

That "P" matrix is the matrix having the eigenvectors corresponding to eigenvalues 4 and -1 so we need to find those eigevectors.

If $\begin{pmatrix} a \\ b \end{pmatrix}$ is an eigen vector corresponding to eigenvalue 4, then we have $\begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}= \begin{pmatrix}a+ 2b \\ 3a+ 2b\end{pmatrix}= \begin{pmatrix}4a \\ 4b\end{pmatrix}$.

So a+ 2b= 4a and 3a+ 2b= 4b. Those reduce to -3a+ 2b= 0 and 3a- 2b= 0 which both reduce to 3a= 2b. There are infinitely many solutions because there are infinitely many eigenvectors corresponding to one eigenvalue (an entire subspace). We can take, as one solution, (a, b)= (2, 3).

If $\begin{pmatrix} a \\ b \end{pmatrix}$ is an eigenvector corresponding to eigenvalue -1, then we have $\begin{pmatrix}1 & 2 \\ 3 & 2\end{pmatrix}\begin{pmatrix}a \\ b\end{pmatrix}= \begin{pmatrix}a+ 2b \\ 3a+ 2b\end{pmatrix}= \begin{pmatrix}-a \\ -b\end{pmatrix}$.

So a+ 2b= -a and 3a+ 2b= -b. Those reduce to 2a+ 2b= 0 and 3a+ 3b= 0 which both reduce to a= -b. There are infinitely many solutions because there are infinitely many eigenvectors corresponding to one eigenvalue (an entire subspace). We can take, as one solution, (a, b)= (1, -1).

So $P= \begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix}$ and $P^{-1}= \begin{pmatrix}\frac{1}{5} & \frac{1}{5} \\ \frac{3}{5} & -\frac{2}{5}\end{pmatrix}$

Now you can check that $P^{1}AP= \begin{pmatrix}\frac{1}{5} & \frac{1}{5} \\ \frac{3}{5} & -\frac{2}{5}\end{pmatrix}\begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}\begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix}= \begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}$

But then $P\begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}P^{-1}= \begin{pmatrix} 1 & 2 \\ 3 & 2\end{pmatrix}$ so we can write the differential equation $\frac{d\begin{pmatrix} y_1 \\ y_2 \end{pmatrix}}{dx}= \begin{pmatrix}1 & 2 \\ 3 & 2 \end{pmatrix}\begin{pmatrix} y_1 \\ y_ 2 \end{pmatrix}$
as
$\frac{d\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}}{dx}= P\begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}P^{-1}\begin{pmatrix} y_1 \\ y_2\end{pmatrix}$

Multiply both sides by $P^{-1}$, which is a constant and can be taken inside the derivative, to get $\frac{dP^{-1}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}}{dx}= \begin{pmatrix} 4 & 0 \\ 0 & -1 \end{pmatrix}P^{-1}\begin{pmatrix}y_1\\ y_2\end{pmatrix}$.

Let $Z= \begin{pmatrix} z_1 \\ z_2\end{pmatrix}= P^{-1}\begin{pmatrix}y_1 \\ y_2\end{pmatrix}$ and the differential equation becomes
$\frac{d\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}}{dx}= \begin{pmatrix}4 & 0 \\ 0 & -1\end{pmatrix}\begin{pmatrix} z_1 \\ z_2 \end{pmatrix}$.

So we have "uncoupled" the differential equations and can write $\frac{dz_1}{dx}= 4z_1$ and $\frac{dz_2}{dx}= -z_2$ which have solutions $z_1= C_1e^{4x}$ and $z_2= C_2e^{-x}$.

All that is left is to go back to $y_1$ and $y_2$. Since $\begin{pmatrix}z_1 \\ z_2\end{pmatrix}= P^{-1}\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}$ we have that $\begin{pmatrix}y_1 \\ y_2 \end{pmatrix}= P\begin{pmatrix}z_1 \\ z_2\end{pmatrix}= \begin{pmatrix}2 & 1 \\ 3 & -1\end{pmatrix}\begin{pmatrix}C_1e^{4x} \\ C_2e^{-x}\end{pmatrix}= \begin{pmatrix}2C_1e^{4x}+ C_2e^{-x} \\ 3C_1e^{4x}- C_2e^{-x}\end{pmatrix}$.

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Country Boy

Well-known member
MHB Math Helper
I thought I would just add that that the two solutions I got in the previous two posts, $y_1= Ae^{-x}+ Be^{4x}$, $y_2= \frac{1}{2}y_1'- \frac{1}{2}y_1=$$-\frac{A}{2}e^{-x}+ 2Be^{4x}- \frac{A}{2}e^{-x}- \frac{B}{2}e^4x=$$ -Ae^{-x}+ \frac{3}{2}Be^{4x}$ and $y_1= 2C_1e^{4x}+ C_2e^{-x}$, $y_2= 3C_1e^{4x}- C_2e^{-x}$, using two different methods, are actually the same.

That is, for any A and B there exist constants, $C_1$ and $C_2$ so that $Ae^{-x}+ Be^{4x}= 2C_1e^{4x}+ C_2e^{-x}$ and $Ae^{-x}+ \frac{3}{2}e^{4x}= 3C_1e^{4x}- C_2e^{-x}$.

If $y_1= Ae^{-x}+ Be^{4x}= 2C_1e^{4x}+ C_2e^{-x}$ for all x then, since $e^{-x}$ and $e^{4x}$ are "independent", We must have $A= C_2$ and $B= 2C_1$. Then $y_2= -Ae^{-x}+ \frac{3}{2}Be^{4x}= -C_2e^{-x}+ 3C_1e^{4x}$ just as we want!

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