# Find the general solution of each of the following systems (using method with constant coefficients).

Find the general solution of each of the following systems (using method with constant coefficients).

I can complete most problems, but there are a few I cannot fully finish.

Problem One:

$$\begin{cases} \frac{dx}{dt}=4x-2y \\ \frac{dy}{dt}=5x-2y \end{cases}\tag{1}$$

$$\left|\begin{pmatrix} 4-r & -2 \\ 5 & 2-r \end{pmatrix}\right| = (4-r)(2-r)+10$$ $$(4-r)(2-r)+10 \iff r=3\pm3i$$ $$\begin{pmatrix} 4-r & -2 \\ 5 & 2-r\end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix}$$

$$=(4-r)A-2B$$ $$=5A+(2-r)B$$

This is as far as I get without the problem becoming very messy..

Problem Two:

$$\begin{cases} \frac{dx}{dt}=5x+4y \\ \frac{dy}{dt}=-x+y \end{cases}\tag{2}$$

$$\left|\begin{pmatrix} 5-r & 4 \\ -1 & 1-r \end{pmatrix} \right|=(5-r)(1-r)+4$$ $$r^2-6r+9=0 \iff r=3$$

$$\begin{pmatrix} 5-r & 4 \\ -1 & 1-r\end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix}$$

$$=(5-r)A+4B$$ $$=(-1)A+(1-r)B$$

Now substitute for when $r=3$:

$$2A+4B=0$$ $$-A-2B=0$$

Using simple algebra we know $A=-2$ and $B=1$. Therefore, $x=-2e^{3t}$, and $y=e^{3t}$. Now we need a second solution of the form $x=(A_1+A_2t)e^t$, and $y=(B_1+B_2t)e^t$. Now I substitute these into our system of differential equation.

$$(A_1+A_2t+A_2)e^t=5(A_1+A_2t)e^t+4(B_1+B_2t)e^t$$ $$(B_1+B_2t+B_2)e^t=-(A_1+A_2t)e^t+(B_1+B_2t)e^t$$

Now, when I try finishing a problem like this one I get something different every time.

Could someone help me finish these two problems, so I can move on and finish the others that are similar to these. Thanks

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Let me know if I got it right; I deduced that you are taking the determinant of a matrix in each problem. We can roll-back to your original post if I messed anything up. – amWhy Nov 11 '12 at 22:31
Ya, that's right. – Purply.Platypus Nov 11 '12 at 22:45
My advice would be to go over whatever notes you have been studying to see how to handle the cases where there are no real eigenvalues and where there is a repeated eigenvlaue. Surely this would have been covered, and formulas derived for these cases. – Gerry Myerson Nov 11 '12 at 23:39

For the first problem, there's no way around working with "messy" computations.

When you write:

$$= (4-r)A - 2B$$ $$= 5A +(2-r)B$$

...what exactly is each expression equal to? Are you simply using '$\;=\;$' to denote first and second entries, respectively, of a $2\times1$ column matrix?

Note that in your second problem there is a problem with the following:

Now substitute for when $r=3$:

$$2A+4B=0$$ $$-A-2B=0$$

Using simple algebra we know $A=-2$ and $B=1$.

If you attempt to solve the system of equations for A and B, if you multiply $(-A-2B=0)$ by $2$ you get $-2A - 4B =0$. Now, adding that equation to the first gives you $0=0$, and hence, there are an infinite number of solutions, $(A, B)$. (The equations are linearly dependent.)

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