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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{equation} \begin{pmatrix} 4-r & -2 \\ 5 & 2-r\end{pmatrix} \begin{pmatrix} A \\ B \end{pmatrix} \end{equation}

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

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

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

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
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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

2 Answers 2

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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(+1) Nicely done Amy. –  B. S. Aug 8 '13 at 11:01

You also can solve the system $$\begin{cases} \frac{dx}{dt}=4x-2y \\ \frac{dy}{dt}=5x-2y \end{cases} $$ as follows. Let $D$ stands for differentiation, so the system will be: $$\begin{cases} Dx=4x-2y \\ Dy=5x-2y \end{cases} \longrightarrow\begin{cases} (D-4)x+2y=0 \\ -5x+(D+2)y=0 \end{cases}$$ Now solve this system by using very elementary way. For example, I multiply the first equation by $5$ and the second by $(D-4)$ and then adding the results, we have: $$(D^2-2D+2)y=0$$ By this way we eliminated the variable $x$. The latter ODE is equivalent to $$y''-2y'+2y=0$$ which can be easily solved. Indeed, we get $$y(t)=C_1e^t\sin t+C_2e^t\cos t$$ Now do the same way for finding an ODE with respect to $x$ and then finding $x(t)$.

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1  
+1 You too! $\quad$ –  amWhy Aug 8 '13 at 11:32
    
How are you, dear friend? ;-) –  amWhy Aug 8 '13 at 12:52
    
@amWhy: Thanks. :) Not bad, just exhausting of the way Life goes ahead. I hope you don't feel like me. :-) –  B. S. Aug 8 '13 at 12:53
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I understand well, what you are feeling. I've "been there" too. Life "cycles"...just ride along until you feel rejuvenated (and you will!) –  amWhy Aug 8 '13 at 12:55

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