91Ó°ÊÓ

First, let's define the given system of equations as: $$ F(x, y) = \begin{bmatrix} f_1(x, y) \\ f_2(x, y) \end{bmatrix} = \begin{bmatrix} \sin x \cos y - 0.1 \\ x^2 + \sin y - 0.2 \end{bmatrix}. $$ Now we need to calculate the Jacobian matrix \(J(x,y)\) of \(F(x,y)\). The Jacobian matrix is: $$ J(x, y) = \begin{bmatrix} \frac{\partial f_1}{\partial x} & \frac{\partial f_1}{\partial y} \\ \frac{\partial f_2}{\partial x} & \frac{\partial f_2}{\partial y} \end{bmatrix}, $$ so we need to find the partial derivatives of the functions \(f_1(x, y)\) and \(f_2(x, y)\): $$ \begin{aligned} \frac{\partial f_1}{\partial x} &= \cos x \cos y \\ \frac{\partial f_1}{\partial y} &= -\sin x \sin y \\ \frac{\partial f_2}{\partial x} &= 2x \\ \frac{\partial f_2}{\partial y} &= \cos y \end{aligned} $$ Putting it all together, we get the Jacobian matrix: $$ J(x, y) = \begin{bmatrix} \cos x \cos y & -\sin x \sin y \\ 2x & \cos y \end{bmatrix} $$

The exercise says we need to find the solution near the point \((x_0, y_0) = (0, 0)\). So let's initialize our first guess as: $$ \begin{bmatrix} x_1 \\ y_1 \end{bmatrix} = \begin{bmatrix} x_0 \\ y_0 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} $$

Now that we have our initial guess, we can apply two iterations of the Newton-Raphson method. The general formula is: $$ \begin{bmatrix} x_{k+1} \\ y_{k+1} \end{bmatrix} = \begin{bmatrix} x_k \\ y_k \end{bmatrix} - J(x_k, y_k)^{-1} \cdot F(x_k, y_k) $$ Now, we apply the first iteration: $$ \begin{bmatrix} x_2 \\ y_2 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} - \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}^{-1} \cdot \begin{bmatrix} -0.1 \\ -0.2 \end{bmatrix} = \begin{bmatrix} 0.1 \\ 0.2 \end{bmatrix} $$ Finally, we apply the second iteration: $$ \Psi=\left\{ \begin{aligned} \frac{d x}{d s}&=F_{u} \\ \frac{d y}{d s}&=F_{v} \\ \frac{d u}{d s}&=F_{p} \\ \frac{d v}{d s}&=F_{q} \end{aligned}\right. \begin{bmatrix} x_3 \\ y_3 \end{bmatrix} = \begin{bmatrix} 0.1 \\ 0.2 \end{bmatrix} - \begin{bmatrix} 0.98007 & -0.19867 \\ 0.2 & 0.98007 \end{bmatrix}^{-1} \cdot \begin{bmatrix} -0.08007 \\ 0.01 \end{bmatrix} \approx \begin{bmatrix} 0.096 \\ 0.192 \end{bmatrix} $$ Thus, after two iterations of the Newton-Raphson method, our approximate solution for the given system of equations is \((x, y) \approx (0.096, 0.192)\).