anonymous
  • anonymous
Please help! I need to find the solution to the system x' = 6y, y' = -5x, with the initial conditions being x(0) = 1, y(0)= 5. I believe I have to do eigenvalues, but my eqns aren't coming out right.
Mathematics
  • Stacey Warren - Expert brainly.com
Hey! We 've verified this expert answer for you, click below to unlock the details :)
SOLVED
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
JamesJ
  • JamesJ
What eigenvalues have you found?
anonymous
  • anonymous
I think I have found lambda equals +/- sqrt(30), based on my matrix: [0- lambda, 6; -5, 0-lambda]
JamesJ
  • JamesJ
+- sqrt(30) i They are imaginary.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
OH. my goodness. Math error! Thanks!
anonymous
  • anonymous
But---
JamesJ
  • JamesJ
and hence the eigenvectors will also be complex. To deal with this, use the same trick you use when the roots of the characteristic equation are imaginary: take sums to find trig functions cos and sin
anonymous
  • anonymous
I want to double check my method: okay, so, I find the values, then the vectors, and then I set up an eqn that looks some thing like x(t) = c1v1e^(eigenvector)t...?
anonymous
  • anonymous
oh not, I replace the e^(eigenvector) with (cost +isint)
JamesJ
  • JamesJ
call the eigenvalues l1 and l2, with corresponding eigenvectors v1 and v2. Then the general solution is c1.v1.e^(l1.t) + c2.v2.e^(l2.t) where c1 and c2 are constants.
anonymous
  • anonymous
but what happens to the (cost + isint)?
JamesJ
  • JamesJ
Yes, change basis and have two new eigenfunctions: 1/2 . ( e^(l1.t) + e^(l2.t) ) = cos(l1.t) 1/2i . ( e^(l1.t) - e^(-l2.t) ) = sin(l1.t) remember that l1 = -l2
amistre64
  • amistre64
never tried these before but i did just watch a video by arthur mattuck tho. x' = 6y +0x y' = 0y - 5y would our matrix be: 6 0 0 -5 or am i going astray there?
JamesJ
  • JamesJ
@amistre: your coefficients are in the wrong place
JamesJ
  • JamesJ
x' = 0x + 6y etc.
amistre64
  • amistre64
we have to start with the x part?
JamesJ
  • JamesJ
The column vector of the functions x and y have to be in the same order on both sides of the equation
anonymous
  • anonymous
yep. And so now I finally have my vector [5; -30i]
anonymous
  • anonymous
nope, i think it should be [5, 30i] for complex conjugates
anonymous
  • anonymous
nvm keep it the way is was
anonymous
  • anonymous
this is wrong...
anonymous
  • anonymous
help :/ I don't know what I am doing any more :C I know that my eigen values are +/- 30i, and when I try to find the eigenvectors, I find -30i*a + 5*b = 0 and -5*a - 30i*b = 0
anonymous
  • anonymous
I don't thing the vector v = [5, -30i] is correct; and this affects my eqns for x and y
amistre64
  • amistre64
\[\begin{vmatrix}6&0\\0&-5\end{vmatrix}\to \lambda^2+\lambda-30=0;\ \lambda={-6,5} \] \[\begin{vmatrix}0&6\\-5&0\end{vmatrix}\to \lambda^2+ 30=0;\ \lambda={\pm i\sqrt{30}} \] so i spose it does matter which column is where
anonymous
  • anonymous
Yeah, the x values are far left and the y values are far right. So your bottom one is correct. Sorry, I didn't put sqrt() around my 30
anonymous
  • anonymous
I am making my vectors again
anonymous
  • anonymous
so it should be v = [-sqrt(30)i; 6]
anonymous
  • anonymous
hello, there anyone else who can help?

Looking for something else?

Not the answer you are looking for? Search for more explanations.