Can You Solve This System of Differential Equations with Initial Values?

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The forum discussion focuses on solving a system of differential equations represented by the equations x' = -x - 4y and y' = 3x - 2y with initial conditions x(0) = 20 and y(0) = 20. Participants discuss the algebraic manipulation of these equations, including the substitution of variables and the differentiation of the first equation to derive a second-order equation x'' - 2x' + 11x = 0. The discussion emphasizes the importance of understanding the underlying concepts from algebra and calculus to effectively solve such differential equations.

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karush
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$\begin{array}{rr}
x' & = -x - 4y\\
y' & = 3x - 2y
\end{array}$
and initial values are
$x(0) = 20\quad y(0) = 20$
so since $x=-x'-4y$
then
$y'=3(-x'-4y)-2y=-3x'-12y-2y=-3x'-14y$
just seeing if this combined eq is ok...i think y also could have been substituted
$4y=-x-x'$ or $y=-\dfrac{x}{4}-\dfrac{x'}{4}$
then
$y'=-3x'+\dfrac{7x}{2}-\dfrac{7x'}{2}$

not that sure :unsure:
 
Last edited:
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The algebra is correct but I don't know why you did that. You have one equation in both x' and y' but that does not help.

I am sure that, in algebra, you learned that to solve two equations in two unknowns you reduce to one equation in one unknown. Here, you can do that by differentiating the first equation again:
x''= -x'- 4y'. The second equation tells us that y'= 3x- 2y so x''= -x- 4(3x- 2y)= -13x+ 8y.

And we can rewrite x'= -x- 4y as 4y= -x'- x so that x''= -13x- 2(-x'- x)= 2x'- 11x or x''- 2x'+11X= 0.

Now solve that for x(t).
 
Last edited:
Country Boy said:
And we can rewrite x'= -x- 4y as 4y= -x'- x so that x''= -13x- 2(-x'- x)= 2x'- 11x or x''- 2x'+11x= 0.
Now solve that for x(t).
are you suggesting to rewrite
$x''- 2x'+11x= 0$ as $r^2-2r+11=0$
thus giving
$r=1+\sqrt{10}i,\quad r=1-\sqrt{10}i$

at lease i saw something like this is some examples
 
You know, from differential Calculus, that the derivative of $e^{rx}$ is $re^{rx}$, a constant times the original function, so it is natural to look for such functions as solutions to equations like this, linear equations with constant coefficients where the various derivatives must cancel.

If $x(t)= e^{rt}$ then $x'(t)= re^{rt}$ and $x''(t)= r^2e^{rt}$ so that $x''- 2x'+ 11X= 0$ becomes $r^2e^{rt}- 2re^{rt}+ 11e^{rt}= (r^2- 2r+11)e^{rt}= 0$. And since $e^{rt}$ is never 0, we must have $r^2- 2r+ 11= 05.

Surely you were taught that in an introductory Differential Equations course?
 
for this topic I have only been able to look at examples in books,...
so I depend heavily on forums for help

why do have a capitol X
 
karush said:
for this topic I have only been able to look at examples in books,...
so I depend heavily on forums for help
I hope you are reading the books to learn the basic concepts and NOT just looking at examples!

why do have a capitol X
The X was a typo. It should have been x''- 2x'+ 11x= 0.
 
both actually but they assume a lot
 
Yes, "Differential Equations" is fairly deep mathematics and requires a lot of prerequisites- largely algebra and Calculus. Have you passed algebra and Calculus courses?
 
yes
i took de
last year but it was all matrix
 

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