System of first order ODE's

Posted on 2007-12-04
Last Modified: 2012-05-05
So a previous question of mine was how to solve the following, by first converting the two first orders into a single second order:

dx/dt = 4x - y
dy/dt = 2x + y + t^2                                  (x(0)=0, y(0)=1)

BigRat pointed out the rather trivial method:

d^2(x)/dt^2 = 4dx/dt - dy/dt
                    = 4 dx/dt - (2x + y + t^2)
                    = 4 dx/dt - 2x - (4x - dx/dt) - t^2
x'' - 5x' +6x = -t^2

Now, solving this, gives:

x = Ae^(2t) + Be^(3t) - 1/6*t^2 - 5/18*t - 19/108       (note: two unknowns)

If I apply x(0)=0:

0 = A + B - 19/108.                 (*)

Now, to solve for y (and substitute in my solution for x(t), from above):

dy/dt = 2x + y + t^2
         = 2(Ae^(2t) + Be^(3t) - 1/6*t^2 - 5/18*t - 19/108) + y + t^2
d/dt(y*e^(-t)) = e^(-t) * {2(Ae^(2t) + Be^(3t) - 1/6*t^2 - 5/18*t - 19/108) + t^2}
y = e^t * Integral { e^(-t) * {2(Ae^(2t) + Be^(3t) - 1/6*t^2 - 5/18*t - 19/108) + t^2} } dt
y = 2Ae^(2t) + Be^(3t) - 1/54*e^(-t)*(23 + 42t + 36t^2) + Ce^t     (3 unknowns)

I can then apply y(0)=1:

1 = 2A + B - 23/54 + C.         (**)

So I have two equations, and three unknowns.... Have I lost information somewhere? Is it possible to eliminate all unknown constants?? I can't see how, but the question seems to imply that this is what is expected.....

Thanks for any input
Question by:InteractiveMind
1 Comment

Accepted Solution

Dimkov earned 500 total points
ID: 20405400
I think you are fine
C can be any number you want to be.
Here is my logic:
dx/dt=... without C

dx/dt will be fine for any set of functions x=... +C

I am not sure about this. It is just an idea

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