numerical-methods

In order to numerically solve the ordinary differential equation dy/dt = -y for t > 0, with an initial condition y(0) = 1, the following scheme is employed (y_{n+1} - y_n) / Δt = -...

In order to numerically solve the ordinary differential equation dy/dt = -y for t > 0 , with an initial condition y(0) = 1 , the following scheme is employed: $\frac{y_{n+1} - y_{n...

The initial value problem $\rm \frac{dy}{dt}+2y=0, y(0)=1$ is solved numerically using the forward Euler’s method with a constant and positive time step of Δt. Let 𝑦 𝑛 represent...

The ordinary differential equation $\frac{dy}{dt} = -\pi y$ subject to an initial condition $y(0) = 1$ is solved numerically using the following scheme: $\frac{y(t_{n+1}) - y(t_n)}...

Find the positive real root of x³ - x - 3 = 0 using Newton-Raphson method. If the starting guess (x₀) is 2, the numerical value of the root after two iterations (x₂) is ________ (r...

Evaluation of $\int_2^4 x^3 dx$ using a 2-equal-segment trapezoidal rule gives a value of _________

The root of the function $$f\left( x \right) = {x^3} + x - 1$$ obtained after first iteration on application of Newton-Raphson scheme using an initial guess of $${x_0} = 1$$ is

Solve the equation $$x = 10\,\cos \,\left( x \right)$$ using the Newton-Raphson method. The initial guess is $$x = {\pi \over 4}.$$ The value of the predicted root after the first...

Gauss-Seidel method is used to solve the following equations (as per the given order). $$${x_1} + 2{x_2} + 3{x_3} = 5$$$ $$$2{x_1} + 3{x_2} + {x_3} = 1$$$ $$$\,3{x_1} + 2{x_2} + {x...

Newton-Raphson method is used to find the roots of the equation, $${\,{x^3} + 2{x^2} + 3x - 1 = 0}$$ If the initial guess is $${x_0} = 1,$$ then the value of $$x$$ after $${2^{nd}}...

Consider an ordinary differential equation $${{dx} \over {dt}} = 4t + 4.\,\,$$ If $$x = {x_0}$$ at $$t=0,$$ the increment in $$x$$ calculated using Runge-Kutta fourth order multi-s...

The real root of the equation $$5x-2cosx=0$$ (up to two decimal accuracy) is

Starting from $$\,{x_0} = 1,\,\,$$ one step of Newton - Raphson method in solving the equation $${x^3} + 3x - 7 = 0$$ gives the next value $${x_1}$$ as

Given the differential equation $${y^1} = x - y$$ with initial condition $$y(0)=0.$$ The value of $$y(0.1)$$ calculated numerically upto the third place of decimal by the $${2^{nd}...