Solving system of coupled differential equations using scipy odeint

Question

I am a bit confused with odeint .

I found one example below to solve y"=ay + by' . So it seems that y[0] is the function, y[1] is the first derivative.

So does the following expression mean y[1] =y' and y'[1]= a*y[0]+b*y[1] ?

If it were y[2], a*y[0]+b*y[1] , what would it mean?

I am a bit confused since the expression does not say the left hand side of the equation.

I also encountered expressions like [a(y[0], y[1]), b(y[0], y[1])] but have no clue of the differential equation.

Here is one example:

from scipy.integrate import odeint
from pylab import * # for plotting commands

def deriv(y,t): # return derivatives of the array y
    a = -2.0
    b = -0.1
    return array([ y[1], a*y[0]+b*y[1] ])

time = linspace(0.0,10.0,1000)
yinit = array([0.0005,0.2]) # initial values
y = odeint(deriv,yinit,time)
figure()
plot(time,y[:,0])
xlabel('t')
ylabel('y')
show()

Answer 1

Let's use Y in deriv instead of y for the rest of answer to be clear:

def deriv(Y,t): # return derivatives of the array Y
    a = -2.0
    b = -0.1
    return array([ Y[1], a*Y[0]+b*Y[1] ])

Function deriv takes Y = [y, y'] as the input.

And it should output their derivatives ( [y', y''] ).

y' = Y[1]

y'' = a*Y[0]+b*Y[1]

Answer 2

Read the documentation on odeint . It requires as an input equation of the following sort:

dy/dt = func(y,t0,...)

As far as I understand it, first element of array([ y[1], a*y[0]+b*y[1] ]) , ie y[1] is put as y in dy/dt which gives dy[1]/dt = y[2] . The second element, ie a*y[0]+b*y[1] serves as func(y,t0,...)