Starting with x = 1, the solution of the equation , after two iterations of Newton-Raphson’s method (up to two decimal places) is ____________

The ordinary differential equation, with  is to be solved using the forward Euler method. The largest time step that can be used to solve the equation without making the numerical solution unstable is ________.

Consider the first order initial value problem , y(0)=1,  with exact solution for x=0.1, the percentage difference between the exact solution and the solution obtained using a single iteration of the second-order Runge-kutta method with step-size h=0. 1is ______________.

The Newton-Raphson method is used to solve the equation. Taking the initial guess as x = 5, the solution obtained at the end of the first iteration is __________.

Match the application to appropriate numerical method.

Application 

P1: Numerical integration

P2: Solution to a transcendental equation

P3: Solution to a system of linear equations

P4: Solution to a differential equation

Numerical Method

M1: Newton-Raphson Method

M2: Runge-Kutta Method

M3: Simpson’s 1/3-rule

M4: Gauss Elimination Method

A polynomial  with all coefficients positive has

A numerical solution of the equation  can be obtained using Newton-Raphson method. If the starting value is x = 2 for the iteration, the value of x that is to be used in the next step is

Consider differential equation  with the initial condition . Using Euler’s first order method with a step size of 0.1, the value of y (0.3) is

The recursion relation to solve  using Newton-Raphson method is

The equation is to be solved using the Newton-Raphson method. If x = 2 is taken as the initial approximation of the solution, then the next approximation using this method will be

Match the following and choose the correct combination:

Simpson’s rule for integration gives exact result when f(x) is a polynomial of degree

Given the differential equation,  with the initial condition . The value of  calculated numerically up to the third place of decimal by the second order Runge Kutta method with step size  is _________