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# System of first order ODE's

Posted on 2007-12-04

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