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" 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?

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