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# Runge-Kutta Method Calculator

To use Runge-Kutta method calculator, enter the equation, write initial condition, interval values, step size, and click calculate button

## Runge-Kutta Method Calculator

Runge Kutta Method Calculator is used to calculate ordinary differential equations (ODEs) numerically.

## What is Runge-kutta method?

**Runge-kutta method **is a numerical technique to find the solution of the ordinary differential equation. The method provides us with the value of **y **as the corresponding value of **x**. It is also known as the **RK method**.

Here we use the most common **RK-4** which is known as runge-kutta of order fourth.

**Formula of Runge-kutta method**

The formula of the Runge-Kutta method of order fourth is given as:

`y`

_{i+1 }= y_{i }+ h/6(k_{1 }+ 2k_{2 }+ 2k_{3 }+ k_{4})

Here,

`x`

_{i+1} = x_{i}+h

where

**i**is the number of iterations.

For the value of **k’s**, we have

`k`

_{1}= f(x_{i}, y_{i})`k`

_{2}= f(x_{i }+ h/2, y_{i }+ hk_{1}/2)`k`

_{3}= f(x_{i }+ h/2, y_{i }+ hk_{2}/2)`k`

_{4}= f(x_{i }+ h, y_{i }+ hk_{3})

## How to calculate Runge-kutta method problems?

Here is a numerical example to understand this iterative method.

**Example:**

If we have differential function **x ^{2}+1 **with the "

**2**" number of iterations. The initial value of the

**x**is

`1 to 3`

and the initial value of **y**is

**2**.

**Solution:**

**Step 1:** The step size is

h= (3-1)/2 = 1, x_{0}= 1, y_{0} = 2 and n= 2

`f(x, y) = x^2+1`

**Step 2: **For **1 ^{st} **iteration

`x`

_{1}= x_{0}+h = 1+1 = 2

k_{1}= f(x_{0}, y_{0})= f(1, 2) = 2

k_{2}= f(x_{0 }+ h/2, y_{0}+ hk_{1}/2) = f(1 + 1/2, 2 + (1)(2)/2) = f(3/2, 3) = 3.25

k_{3}= f(x_{0 }+ h/2, y_{0 }+ hk_{2}/2) = f(1 + 1/2, 2 + (1)(3.25)/2) = f(2/2, 3.625) = 3.25

k_{4} = f(x_{0 }+ h,y_{0 }+ hk_{3}) = f(1 + 1, 2 + (1)(3.25)) = f(2, 5.25) = 5

y_{1} = y(x_{1}) = y(2) = y_{0}+ h/6(k_{1}+2k_{2}+2k_{3}+k_{4}) = 2+1/6(2+2(3.25)+2(3.25)+5)

`y`

_{1} = 5.333

**Step 3: **For the **2 ^{nd}** iteration we have

**x**and

_{1}**y**to calculate

_{1}**x**and

_{2}**y**.

_{2}`x`

_{2}= x_{1}+h = 2+1 = 3

k_{1}= f(x_{1}, y_{1}) = f(2, 5.333) = 5

k_{2}= f(x_{1}+ h/2, y_{1}+ hk_{1}/2) = f(2 + 1/2, 5.333 + (1)(5)/2) = f(5/2, 7.83) = 7.25

k_{3}= f(x_{1}+ h/2, y_{1}+ hk_{2}/2) = f(2 + 1/2, 5.333 + (1)(7.25)/2= f(5/2, 8.958) = 7.25

k_{4} = f(x_{1 }+ h, y_{1 }+ hk_{3}) = f(2+1, 5.333 + (1)(7.25)) = f(3, 12.583) = 10

y_{2} = y(x_{2}) = y (3) = y_{1}+ h/6(k_{1}+ 2k_{2 }+ 2k_{3 }+ k_{4}) = 5.333 + 1/6(5 + 2(7.25) + 2(7.25) +10)

`y`

_{2} = 12.666

Hence the approximate value of** y _{2} = 12.666**.