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

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Answer 1 · 2016-09-10T14:22:06

See below.

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.

How do you solve the system 4xy + 4y^2 - 3cos(y) = -24xy+4y2−3cos(y)=−24xy + 4y^2 - 3cos(y) = -2, 5x^2 - 2xy^2 - 5sin(x) = -55x2−2xy2−5sin(x)=−55x^2 - 2xy^2 - 5sin(x) = -5?