" Solve the system of equations
$x-2y+y^{2}+y^{3}-4 = 0$ and $-x-y+2y^{2}-1 = 0$
starting with the point (0,0)."
First I have found a homotopy.
h(t,k) = f(k) + (t-1)f($k_{0}$) where $k_{0} = (0,0)$
$h(t,k) = \left( \begin{array}{c} x-2y+y^{2}+y^{3}-4t \\ -x-y+2y^{2} - t \\ \end{array} \right) $
Now I define a vector v = (t,x,y) where x, y is the components of k. I also assume that t, x, y is a function of an independent variable z.
So I have:
h'(v(z))v'(z)=0
I have found h'(v(z)) to be:
h'(v(z))=\left( \begin{array}{ccc} -4 & 1 & -2+2y+3y^{2} \\ -1 & -1 & -1+4y \\ \end{array} \right)
Solving the above differential equation:
t'(z) = -6y+3-3y^{2}
x'(z) = 2 - 14y + 3y^{2}
y'(z) = -5
Now I want to find t(z), x(z) and y(z). I integrate and I know that t(0) = 0, x(0) = 0 , y(0) = 0.
$t(z) = -6yz+3z-3y^{2}z$
$x(z) = 2z - 14yz + 3y^{2}z$
$y(z) = -5z$
Which also means when I substitue y(z) in t(z) and x(z):
$t(z) = 30z^{2}+3z-75z^{3}$
$x(z) = 2z +70z^{2} + 75z^{3}$
$y(z) = -5z$
Now I want to find the z that makes t(z)=1 true.
I solve this equation and get:
z=-.1883278521, z=.1687291036, z=.4195987485
Now i find x(-.1883278521)=1.605098728, y(-.1883278521) =0.9416392605
And I check if the values solve the original equation. unfortunately they dont, and I really want to figure out what I have done wrong. Someone who can give a hint?