Solve the following systemusing the Newton-Raphsonmethod for one iteration,given that x(0)=y(0)=0.5:f(x) = 3yx² - 10x + 7g(x) = y² - 5y +4O x = 0.8180 and y =0.9375O x = 0.8181 and y =0.9357O x = 0.818 and y =-0.9375O x = -0.8181 and y =0.9375

We will use the Newton Raphson method of dual variable to solve the given system…

Answered: Solve the following system using the…

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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A 100-L tank was initially empty, but is to be filled with salt-and-water solution. The way the solution is being poured into the tank is through a device the controls the rate at which it is being poured. The controller ensures the solution is being poured at time t (minutes) with the expression (1 + cos(2t)) in liters per minute. It follows that the amount of salt for every liter of the solution being poured at time t (minutes) is described by the expression 0.2(1 + cos(2t)) in kilograms. This implies the rate the salt- and-water solution being poured into the tank alternately gets faster and slower. Determine which of the following differential equation describing the salt concentration S in the solution at any time t. O A. dS dt OB. dS dt OC. dS dt O D. dS dt = 0.2(1 + cos 21) ² = = = -0.2(1 + cos 21) ² 0.2 (1 + cos 21) ² 0.2 2 (1 + cos 21) ²
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Given the function : f(x) = x³ + 5x² + 7x-20; solve using Secant Method with initial values of (1.000, 2.000) 1st iteration X0 = 1.000 X1 2.000 xnew = 1.241 2nd iteration X0 = 2.000 X1 2.000 xnew 1.296 3rd iteration X0 = 1.241 X1 = 1.296 xnew= 1.310 4th iteration X0 = 1.296 X1 = 1.310 xnew= 1.310 5th iteration X0 = -0.001 X1 = -0.001 xnew= 1.310 f(x0) = -7.000 f(x1) = 22.000 f(xnew) = -1.701 f(x0) = 22.000 f(x1) = -1.701 f(xnew) = -0.353 f(x0) = -1.701 f(x1) = -0.353 f(xnew) -0.001 = f(x0) = -0.001 f(x1) = -0.001 f(xnew) = -0.001 f(x0) = 1.310 f(x1) = -0.001 f(xnew) = -0.001
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Solve the following system using the Newton-Raphson method for one iteration, given that x(0)=y(0)=0.5: f(x)= 3yx² - 10x + 7 g(x)= y² - 5y +4 O x = 0.8180 and y =0.9375 O x = 0.8181 and y =0.9357 O x = 0.818 and y =-0.9375 O x = -0.8181 and y =0.9375