Question

Q5) consider the nonlinear system x3 - 3xy2 - 2x +2 = 0 3x2y - y3 - 2y = 0 Find x and y by Newton-Raphson method if x =1, x2 =1.

Answer 1

A system of non - linear equations takes a lot of time to solve by hand but can be solved easily by using MATLAB

I explained the solution detailedly by hand and also provided the MATLAB code for your understanding

This is a lengthy answer. I'm hoping that you have some idea about MATLAB. If you don't, don't worry. I explained everything in detail.

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equations is Griven system of non-linear x - 3xy" - 2x + 2 = 0 3x²y -ye - 24 [2] [:] Newton - Raphson method for system of equations is given as (n+1) x (n) [Jen] [ f cx (nl) Initial guess xº non-linear where J is a Jacobian matrix For a 2x2 matrix, af, (x, y) of Cary J= 2x dy a faca,y) afz(x, y) ay let ficary) x² 3xy 3xy - 2x +2=0 falary) 3x²2y - y - 2y=0 Partial differentiations afi 3x² By - 2 ay 3x²3y - 2 afz Gry dy ; afe = -6xy 2x 2 afz za

JE 3x² By²-2 - 6xy 3 x ² 34 22 6xy Iteration 1 - JO) = J at x i. e Jat x, y ho have xM+) _ £12 [Jen] If (200)] n=0 = x ² 2 2 [200] [f(x)] [:] [3 (1) ²3 (1) 2-2 3-30022 [ - (0) 2 -6 CIOCO 6 CIOCI -6 6 2 2 f (x con (0) x - [=cos] [f Ce co) [] -2 -6 - 2 0 6 - 2 0.9 O. Z

sterations, - (2) (1) x x [JC10] [f (x ) -0.394 C-l-oy -3.78 - 0.88 37 0.6003 steration 3 : (3) x x (2) - [5 (27] [ f cx (2)] Color 0.8837 -0.7383 -3.1829 F0.0326 Or6003 - 3.1829 -0.0106 - 0.7383 0.8846 0.58 98 steration 4: (4) Similarly, we get 0.88 46 0.5897 values 0.5897 0.8846 After ateration 5, x(5). L are converged (x, y) = (0.8846, 0.5897)
In the above images, inverse of J is calculated in MATLAB. Type inv(J) to get inverse matrix

MATLAB CODE

Editor Window: Type the below code in editor window. Save this code as "NR.m" file for the code to work

function [f,J,inverseJ,x0,xnew]= NR(x,y)
x0=[x;y];
f1 = x^3 - 3*x*y^2 - 2*x + 2;
f2 = 3*x^2*y - y^3 - 2*y;
f=[f1;f2];
J = [3*x^2 - 3*y^2 - 2 -6*x*y; 6*x*y 3*x^2 - 3*y^2 - 2];
inverseJ = inv(J);
xnew = x0 - inverseJ*f;

Iteration 1

Type this in command window and click enter

[f,J,inverseJ,x0,xnew]= NR(1,1)

Output

This will be the output displayed:

f =
-2
0
J =
-2 -6
6 -2
inverseJ =
-0.0500 0.1500
-0.1500 -0.0500
x0 =
1
1
xnew =
0.9000
0.7000

Now again call your function by entering these xnew values as shown below

Iteration 2

Type this in command window

[f,J,inverseJ,x0,xnew]= NR(0.9,0.7)

Output

This will be the output displayed:

f =
-0.3940
-0.0420
J =
-1.0400 -3.7800
3.7800 -1.0400
inverseJ =
-0.0677 0.2459
-0.2459 -0.0677
x0 =
0.9000
0.7000
xnew =
0.8837
0.6003

Similarly, if you repeat the above process for another 3 times, after 5th iteration the values will be converged and required x,y values are obtained as shown below:

>> [f,J, inverseJ, x0, xnew]= NR (0.8846,0.5897) f = 1.0e-03 * 0.1661 -0.1148 = -3.1299 -0.6957 -0.6957 3.1299 inverseJ -0.0677 -0.3045 х0 0.8846 0.5897 0.3045 -0.0677 = xnew 0.8846 0.5897

Therefore, the required values are (x, y)= (0.8846, 0.5897)