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

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In summary: I am an AI and unable to understand your frustration. Is there something specific you need help with?
  • #1
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:
 
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  • #2
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).
 
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  • #3
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
 
  • #4
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?
 
  • #5
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
 
  • #6
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.
 
  • #7
both actually but they assume a lot
 
  • #8
Yes, "Differential Equations" is fairly deep mathematics and requires a lot of prerequisites- largely algebra and Calculus. Have you passed algebra and Calculus courses?
 
  • #9
yes
i took de
last year but it was all matrix
 

1. What is a system of equations?

A system of equations is a set of two or more equations with multiple variables that have a common solution. This means that the values of the variables will satisfy each equation in the system.

2. How do you solve a system of equations?

There are several methods for solving a system of equations, including substitution, elimination, and graphing. The method used depends on the type and complexity of the equations in the system.

3. What is the purpose of solving a system of equations?

Solving a system of equations allows us to find the values of the variables that satisfy all of the equations in the system. This can be useful in many real-world applications, such as finding the intersection point of two lines or determining the optimal solution to a problem.

4. Can a system of equations have more than one solution?

Yes, a system of equations can have no solution, one solution, or infinitely many solutions. The number of solutions depends on the relationship between the equations in the system.

5. What is the difference between consistent and inconsistent systems of equations?

A consistent system of equations has at least one solution, while an inconsistent system has no solutions. This means that the equations in a consistent system are not contradictory, while the equations in an inconsistent system are contradictory and cannot be satisfied simultaneously.

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