# Can anyone help me solve this problem using Newton's Method?

2 views (last 30 days)
Muzammil Arif on 13 Mar 2021
Commented: Sergey Kasyanov on 15 Mar 2021
Using Newton's Method solve the equations:
x^3 + y^2 + x - 0.715 = 0
x^2 + y^3 - y - 0.523 = 0
with (x0, y0) = (1, 0).

Sergey Kasyanov on 13 Mar 2021
Hello,
try that
i = 1;
V0 = [1;0];%[x0;y0]
V = V0 + 1;
while max(abs(V0 - V)) > 1e-10 && i < 1e2
V = V0;
J = [3*V(1)^2+1 2*V(2)
2*V(1) 3*V(2)^2-1];
F = [V(1)^3+V(2)^2+V(1)-0.715
V(1)^2+V(2)^3-V(2)-0.523];
V0 = V - F./(J*V);
i = i + 1;
end
x = V(1);
y = V(2);
Sergey Kasyanov on 15 Mar 2021
There are some solutions of the equations system. You can try to find another solutions by changing start point or by adding and varying coefficient k in equation (0<k<1):
V0 = V - k * F./(J*V);
There are three solution near the 0: [-0.46, 1.13], [0.38;-0.54], [0.5;-0.3].

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