Solve by the Newton–Raphson method the simultaneous, nonlinear equationsfor x1 and x2 to within ±0.001. As initial guesses, assume:(a) x1 =0.4, x2 = 0.9; (b) x1 =0.6, x2 = 0.9; (c) x1 = 1.0, x2 = 1.0; sin(TX,2)-꼭-x=0exp(2x1)1- 2x1 +x24т4т+ exp(1) I, vMESHT, V(loop)MESHMESH(inner loop)equationsequationsequations(loop)x, y, T,V, LI, v,K, hparametersK,K,hy, hLhy, hтComplexthermodynamicmodelsComplexthermodynamicmodelsX, y, T,V, L.(outer loop)ApproximateRect thermodynamicmodelsK,hy, h.(a)(ь)Complexthermodynamicmodels(c)

Answered: sin(TX,2)-꼭-x=0 exp(2x1) 1- 2x1 +x2 4т…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Solve by the Newton–Raphson method the simultaneous, nonlinear equations
for x₁ and x₂ to within ±0.001. As initial guesses, assume:
(a) x₁ =0.4, x₂ = 0.9; (b) x₁ =0.6, x₂ = 0.9; (c) x₁ = 1.0, x₂ = 1.0;

sin(TX,2)-꼭-x=0
exp(2x1)
1- 2x1 +x2
4т
4т
+ exp(1)

I, v
MESH
T, V
(loop)
MESH
MESH
(inner loop)
equations
equations
equations
(loop)
x, y, T,
V, L
I, v,
K, h
parameters
K,
K,
hy, hL
hy, h
т
Complex
thermodynamic
models
Complex
thermodynamic
models
X, y, T,
V, L.
(outer loop)
Approximate
Rect thermodynamic
models
K,
hy, h.
(a)
(ь)
Complex
thermodynamic
models
(c)

Expert Solution

Introduction

As per the question we are given a system of nonlinear equations in variables x₁ → x , x₂ → y

as :

And we have to use the Newton-Raphson method to find the solution with tolerance ±0.001

using the following initial guesses :

x₁ = x = 0.4 , x₂ = y = 0.9
x₁ = x = 0.6 , x₂ = y = 0.9
x₁ = x = 1.0 , x₂ = y = 1.0

Solution (a)

So the approximate root using Newton Raphson method becomes :

x₁ = x = 0.2994 , x₂ = y = 0.903

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