is it posible to solve a non-liner system...and how!!!
for example
$\displaystyle x^2 +y^3 =10 $
$\displaystyle x^3 -y^4 =-6 $
if any one has any thoughts on how you do this please let me know
~dan
Of course,Originally Posted by dan
$\displaystyle x^3=y^4-6$
$\displaystyle x=\sqrt[3]{y^4-6}$
Substitute,
$\displaystyle (y^4-6)^{2/3}+y^3=10$
Thus,
$\displaystyle (y^4-6)^{2/3}=(10+y^3)$
Cube both sides,
$\displaystyle (y^2-6)^2=(10+y^3)^3$
Expand and solve for the nasty nontic.
Note, the solutions not necessarily can be expressed in algebraic terms.
using:Originally Posted by ThePerfectHacker
$\displaystyle x=\pm \sqrt{10-y^3}$
and substituting into
$\displaystyle
x^3 -y^4 =-6
$
will give two octics - a much easier problem than a single nonic. ?
RonL
If you are happy with approximate solutions one could start garphically.Originally Posted by dan
These two equations represent curves in the x-y plane. Sketch them
and your first approximation to the soution/s will be the point/s of
intersection.
Then they can be refined with an appropriate numerical procedure (say
linearising about the approximate roots, or something like that).
RonL
Not necessarily there is a general method.Originally Posted by dan
Yes, you do expand with binomial theorem then work with polynomial equation.
(Do you not just love it when mathematicians say "I cannot solve the problem but I can show it cannot be solved!")
so if i had a polynomal. could i solve with newton??
i dont so how you get a nontic though $\displaystyle (10 +y^3)^3$ just because 3*3 is nine???
some how i got from $\displaystyle (y^2 -6)^2 =(10 +y^3)^3 $
to $\displaystyle 0= y^5 + 59y^4 +90y^3 + 12y^2 + 964 $
is that right???
dan
What do you get if you substitute this back into the original equations?Originally Posted by dan
Is there a x which with this as a value of y solves the equations?
RonL
(when I do this I get no consistent solution for x, but I could be doing the
arithmetic wrongly )