How do you solve the system 4xy + 4y^2 - 3cos(y) = -2, 5x^2 - 2xy^2 - 5sin(x) = -5?

1 Answer
Sep 10, 2016

See below.

Explanation:

Using an iterative process.

Making f(x,y) = ((4xy+4y^2-3cos(y)+2),(x^2-2xy^2-5sin(x)+5))

Calling p = (x,y) and p_0 = (x_0,y_0)

The linear expansion around x_0,y_0 is

f(p) approx f(p_0)+grad f_(p_0)(p-p_0)

Supposing that abs(f(p)) approx 0 we have the iterative procedure

p = p_0 - (grad f_(p_0))^(-1)f(p_0) or

p_(k+1) = p_k - (grad f_(p_k))^(-1)f(p_k)

Here

grad f = ((4 y, 4 x + 8 y + 3 sin(y)),(2 x - 2 y^2 - 5 cos(x), -4 x y))

Starting with p_0 = (2.32703,-2.11049) we obtain the convergent sequence

((1.70737,-1.42624), (1.15774,-1.01564), (1.00499,-0.939189), (1.00591,-0.942905), (1.00591,-0.942895))

within 6 significant digits.