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

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Solution

Solution provided by AtoZmath.com

1. Find y(1) for

y′′=-4z-4y $y''=-4z-4y$ ,

x0=0,y0=0,z0=1 $x_0=0, y_0=0, z_0=1$ , with step length 0.1

2. Find y(0.1) for

y′′=1+2xy-x2z $y''=1+2xy-x^2z$ ,

x0=0,y0=1,z0=0 $x_0=0, y_0=1, z_0=0$ , with step length 0.1

3. Find y(0.2) for

y′′=xz2-y2 $y''=xz^2-y^2$ ,

x0=0,y0=1,z0=0 $x_0=0, y_0=1, z_0=0$ , with step length 0.2

Example

1. Find y(0.1) for y′′=1+2xy-x2z $y''=1+2xy-x^2z$ , x0=0,y0=1,z0=0 $x_0=0, y_0=1, z_0=0$ , with step length 0.1 using Runge-Kutta 4 method (2nd order derivative)

Solution:
Given

y′′=1+2xy-x2z,y(0)=1,y′(0)=0,h=0.1,y(0.1)=? $y^('')=1+2xy-x^2z, y(0)=1, y'(0)=0, h=0.1, y(0.1)=?$

put

dydx=z $(dy)/(dx)=z$ and differentiate w.r.t. x, we obtain

d2ydx2=dzdx $(d^2y)/(dx^2)=(dz)/(dx)$

We have system of equations

dydx=z=f(x,y,z) $(dy)/(dx)=z=f(x,y,z)$

dzdx=1+2xy-x2z=g(x,y,z) $(dz)/(dx)=1+2xy-x^2z=g(x,y,z)$

Forth order R-K method for second order differential equation

k1=hf(x0,y0,z0)=(0.1)⋅f(0,1,0)=(0.1)⋅(0)=0 $k_1=hf(x_0,y_0,z_0)=(0.1)*f(0,1,0)=(0.1)*(0)=0$

l1=hg(x0,y0,z0)=(0.1)⋅g(0,1,0)=(0.1)⋅(1)=0.1 $l_1=hg(x_0,y_0,z_0)=(0.1)*g(0,1,0)=(0.1)*(1)=0.1$

k2=hf(x0+h2,y0+k12,z0+l12)=(0.1)⋅f(0.05,1,0.05)=(0.1)⋅(0.05)=0.005 $k_2=hf(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*f(0.05,1,0.05)=(0.1)*(0.05)=0.005$

l2=hg(x0+h2,y0+k12,z0+l12)=(0.1)⋅g(0.05,1,0.05)=(0.1)⋅(1.09988)=0.10999 $l_2=hg(x_0+h/2,y_0+k_1/2,z_0+l_1/2)=(0.1)*g(0.05,1,0.05)=(0.1)*(1.09988)=0.10999$

k3=hf(x0+h2,y0+k22,z0+l22)=(0.1)⋅f(0.05,1.0025,0.05499)=(0.1)⋅(0.05499)=0.0055 $k_3=hf(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.1)*f(0.05,1.0025,0.05499)=(0.1)*(0.05499)=0.0055$

l3=hg(x0+h2,y0+k22,z0+l22)=(0.1)⋅g(0.05,1.0025,0.05499)=(0.1)⋅(1.10011)=0.11001 $l_3=hg(x_0+h/2,y_0+k_2/2,z_0+l_2/2)=(0.1)*g(0.05,1.0025,0.05499)=(0.1)*(1.10011)=0.11001$

k4=hf(x0+h,y0+k3,z0+l3)=(0.1)⋅f(0.1,1.0055,0.11001)=(0.1)⋅(0.11001)=0.011 $k_4=hf(x_0+h,y_0+k_3,z_0+l_3)=(0.1)*f(0.1,1.0055,0.11001)=(0.1)*(0.11001)=0.011$

l4=hg(x0+h,y0+k3,z0+l3)=(0.1)⋅g(0.1,1.0055,0.11001)=(0.1)⋅(1.2)=0.12 $l_4=hg(x_0+h,y_0+k_3,z_0+l_3)=(0.1)*g(0.1,1.0055,0.11001)=(0.1)*(1.2)=0.12$

Now,

y1=y0+16(k1+2k2+2k3+k4) $y_1=y_0+1/6(k_1+2k_2+2k_3+k_4)$

y1=1+16[0+2(0.005)+2(0.0055)+(0.011)] $y_1=1+1/6[0+2(0.005)+2(0.0055)+(0.011)]$

y1=1.00533 $y_1=1.00533$

∴y(0.1)=1.00533 $:.y(0.1)=1.00533$

Input functions

Sr No.	Function	Input value
1.	x3 $x^3$	x^3
2.	√x $sqrt(x)$	sqrt(x)
3.	3√x $root(3)(x)$	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) $sin^(-1)(x)$	asin(x)
11.	cos-1(x) $cos^(-1)(x)$	acos(x)
12.	tan-1(x) $tan^(-1)(x)$	atan(x)
13.	sin2(x) $sin^2(x)$	sin^2(x)
14.	logy(x) $log_y(x)$	log(y,x)
15.	log10(x) $log_10(x)$	log(x)
16.	loge(x) $log_e(x)$	ln(x)
17.	ex $e^x$	exp(x) or e^x
18.	e2x $e^(2x)$	exp(2x) or e^(2x)
19.	∞ $oo$	inf