Question

Solving simultaneous differential equation?
Solving simultaneous differential equation

dx /dt = - xA / (x+y)
dy /dt = -y B / ( x+ y )
A and B are constants x and y are functions of t . I want to know value of x and y in terms of constants and t . Please help and suggest literature.

What should be matrix if I have to solve by eigen value method , what should be appoach please help .

Answer 1

elementary differential equations by Boyce DiPrima
To solve this things, i'll use Cayley-Hamilton theorem
I don't know if you use numeric methods to solve this stuff or not so...Read xD

Answer 2

But how to put into matrix form . As for following diff eqn form
dx/dt = A x + B y
dy / dt = Cx + D y

matrix becomes
A B
C D
eigenvalues and eigen vectors are calculated .
What should be matrix in my my case as I have diff eqn
dx /dt = - xA / (x+y)
dy /dt = -y B / ( x+ y )
Please put light .

Answer 3

i don't think you can put it in matrix form because it is Non-linear
your example is system of linear diff equations....the given system is non-linear so other methods must be used
-I don't think it has an explicit solution for x(t) and y(t)
you can make the inference that:
\[\frac{dy}{dx} = \frac{dy}{dt}*\frac{dt}{dx}\]
\[\frac{dy}{dx} = \frac{By}{Ax}\]
thus
\[y = x^{B/A}\]
then by substitution
\[\frac{dx}{dt} = \frac{-Ax}{x+x^{B/A}}\]
simplify
\[\frac{dx}{dt} = \frac{-A}{1+x^{B/A -1}}\]
thus
\[x + \frac{A}{B}x^{B/A} = -At\]
(ignoring constants of integration)
from here you are stuck isolating x(t) ?

hope this helps
here is wolfram solution
http://www.wolframalpha.com/input/?i=x%27%28t%29+%3D+-Ax%2F%28x%2By%29+%2C+y%27%28t%29+%3D+-By%2F%28x%2By%29

Answer 4

Thanks both Dumbcow and Umangiasd.

Answer 5

In the step
\[ \frac{ dy }{ dx } = \frac{ By }{ Ax } ,y = x^{\frac{ b }{ a }} \]
or
\[y = k(t)x ^{\frac{ b }{ a }}\] , k is substituted as a constant having logarithmic form .
Will there be integration constant . I think that constant may be a function of t . Am I right or not .

Answer 6

@ankushsci
yes you are right, i ignored the integration constant there, however it is just a constant NOT a function of t

Answer 7

Thanks dumbcow.

Answer 8

yw