Solve numerical differential equation using Runge-Kutta 4 method (1st order derivative) calculator

SolutionHelpInput functions

Solution

Solution provided by AtoZmath.com

1. Find y(0.5) for `y'=-2x-y`, y(0) = -1, with step length 0.1

2. Find y(2) for `y'=(x-y)/2`, y(0) = 1, with step length 0.2

3. Find y(0.3) for `y'=-(x*y^2+y)`, y(0) = 1, with step length 0.1

4. Find y(0.2) for `y'=-y`, y(0) = 1, with step length 0.1

Example

1. Find y(0.2) for `y'=(x-y)/2`, `x_0=0, y_0=1`, with step length 0.1 using Runge-Kutta 4 method (1st order derivative)

Solution:
Given `y'=(x-y)/2, y(0)=1, h=0.1, y(0.2)=?`

Forth order R-K method
`k_1=hf(x_0,y_0)=(0.1)f(0,1)=(0.1)*(-0.5)=-0.05`

`k_2=hf(x_0+h/2,y_0+k_1/2)=(0.1)f(0.05,0.975)=(0.1)*(-0.4625)=-0.04625`

`k_3=hf(x_0+h/2,y_0+k_2/2)=(0.1)f(0.05,0.97688)=(0.1)*(-0.46344)=-0.04634`

`k_4=hf(x_0+h,y_0+k_3)=(0.1)f(0.1,0.95366)=(0.1)*(-0.42683)=-0.04268`

`y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)`

`y_1=1+1/6[-0.05+2(-0.04625)+2(-0.04634)+(-0.04268)]`

`y_1=0.95369`

`:.y(0.1)=0.95369`

Again taking `(x_1,y_1)` in place of `(x_0,y_0)` and repeat the process

`k_1=hf(x_1,y_1)=(0.1)f(0.1,0.95369)=(0.1)*(-0.42684)=-0.04268`

`k_2=hf(x_1+h/2,y_1+k_1/2)=(0.1)f(0.15,0.93235)=(0.1)*(-0.39117)=-0.03912`

`k_3=hf(x_1+h/2,y_1+k_2/2)=(0.1)f(0.15,0.93413)=(0.1)*(-0.39206)=-0.03921`

`k_4=hf(x_1+h,y_1+k_3)=(0.1)f(0.2,0.91448)=(0.1)*(-0.35724)=-0.03572`

`y_2=y_1+1/6(k_1+2k_2+2k_3+k_4)`

`y_2=0.95369+1/6[-0.04268+2(-0.03912)+2(-0.03921)+(-0.03572)]`

`y_2=0.91451`

`:.y(0.2)=0.91451`

`:.y(0.2)=0.91451`

Input functions

Sr No.	Function	Input value
1.	`x^3`	x^3
2.	`sqrt(x)`	sqrt(x)
3.	`root(3)(x)`	root(3,x)
4.	sin(x)	sin(x)
5.	cos(x)	cos(x)
6.	tan(x)	tan(x)
7.	sec(x)	sec(x)
8.	cosec(x)	csc(x)
9.	cot(x)	cot(x)
10.	`sin^(-1)(x)`	asin(x)
11.	`cos^(-1)(x)`	acos(x)
12.	`tan^(-1)(x)`	atan(x)
13.	`sin^2(x)`	sin^2(x)
14.	`log_y(x)`	log(y,x)
15.	`log_10(x)`	log(x)
16.	`log_e(x)`	ln(x)
17.	`e^x`	exp(x) or e^x
18.	`e^(2x)`	exp(2x) or e^(2x)
19.	`oo`	inf