Question

Please write simple python code of y''+2y'-3y = 0, y(0) = 2, y'(0)=1 with modified euler method and 4th order Runge-Kutta Kutta method

Answer 1

'''
y'' + 2y' -3y = 0, y(0) = 2, y'(0)=1
let y' = z
giving 2 ODEs to solve simulteneously
z' + 2z - 3y = 0 and y' = z for y(0) = 2, z(0) = 1

assuming a step size h and time vector tmin<=t<=tmax
'''

import matplotlib.pyplot as plt1;import matplotlib.pyplot as plt2

# function to form and return the two 1st-order ODE simulteneous equations
def f(y,z):
dydt = z;
dzdt = 3*y - 2*z;

return dydt,dzdt

# funcion to map the output of funtion f() to only y
def fy(y,z):
[dydt,dzdt] = f(y,z)
return dydt

# funcion to map the output of funtion f() to only z
def fz(y,z):
[dydt,dzdt] = f(y,z)
return dzdt   

# setting the necessary constants
h = 0.1
tmin = 0; tmax = 1;
n = int((tmax-tmin)/h);
y0 = 2; z0 = 1;

# ---------- SOLVING THE ODEs SIMULTENEOUSLY --------

# modified euler

# setting initial conditions
t = [];y=[];z=[]
t.append(tmin);
y.append(y0);
z.append(z0);

for i in range(n):
'''
# ensure y arguments are always on the left, z arguments on the right of either fy or fz
# the x or t components of modified euler method are left out since they do not originally appear in the ODE
'''
ytemp = y[i] + (h/2)*( fy(y[i],z[i]) + fy( y[i]+ h*fy(y[i],z[i]), z[i]+h*fz(y[i],z[i]) ))
ztemp = z[i] + (h/2)*( fz(y[i],z[i]) + fz( y[i]+ h*fy(y[i],z[i]), z[i]+h*fz(y[i],z[i]) ))

# update the y and z vectors
y.append(ytemp)
z.append(ztemp)

# update the time vector
ttemp = t[i]+ h;
t.append(ttemp)

plt1.figure(1)
plt1.plot(t,y,'g-')
#plt1.plot(t,z,'r-')
plt1.xlabel('t');plt1.ylabel('y(t)'),plt1.title('Modified Euler')
plt1.grid()
#------------------------------------------------------------------

# 4th order Runge-Kutta Kutta method

# setting initial conditions
y = [];z = []; t=[]
t.append(tmin);
y.append(y0);
z.append(z0);

for i in range(n):
'''
# the K's are identified with y or z for each cartegory
# ensure y arguments are always on the left, z arguments on the right of either fy or fz
# the x or t components of 4th order Runge-Kutta Kutta method are left out since they do not originally appear in the ODE
'''
yK1 = h*fy(y[i],z[i])
zK1 = h*fz(y[i],z[i])

yK2 = h*fy(y[i]+yK1/2,z[i]+zK1/2)
zK2 = h*fz(y[i]+yK1/2,z[i]+zK1/2)

yK3 = h*fy(y[i]+yK2/2, z[i]+zK2/2)
zK3 = h*fz(y[i]+yK2/2, z[i]+zK2/2)

yK4 = h*fy(y[i]+yK3,z[i]+zK3)
zK4 = h*fz(y[i]+yK3,z[i]+zK3)

ytemp = y[i] + yK1/6 + yK2/3 + yK3/3 + yK4/6
ztemp = z[i] + zK1/6 + zK2/3 + zK3/3 + zK4/6

  
# update the y and z vectors
y.append(ytemp)
z.append(ztemp)

# update the time vector
ttemp = t[i]+ h;
t.append(ttemp)

plt2.figure(2)
plt2.plot(t,y,'g-')
#plt2.plot(t,z,'r-')
plt2.xlabel('t');plt2.ylabel('y(t)'),plt2.title('4th order Runge-Kutta Kutta')
plt2.grid()
plt1.show();plt2.show()

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