Determine coefficients of a differential equation

[jackonelli]

Hi there.

I've been struggling with this problem for days now (4 days, no joke) and I feel like I have a mental block and really cannot get any further.

I have a system that's described by

f(t) = g''(t) + 15g'(t) + 1600g(t) Where the input is g(t)
The problem is to, with this information, determine the coefficients in another system, where the input isf(t)and the output is given by u(t), so that u(t) = g(t)
This other system is given by
c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0f(t)

I think this is supposed to be simple and I think I make it more difficult in my head than it is. I first substituted f(t) in the second differential equation with the left hand side of the first equation:

c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0(g''(t) + 15g'(t) + 1600g(t))
=> c_2u''(t)+c_1u'(t)+c_0u(t) = b_2f''(t) + b_1f'(t) + b_0g''(t) + b_015g'(t) + b_01600g(t)I´m supposed to get numerical values for all coefficients but I really can’t figure out what coefficients makes u(x) = g(t). I don't really know how to proceed. I've tried a lot of other ways too, for example solve for u(t) and g(t) explicitly:u(t) = \frac{b_2}{c_0}f''(t) + \frac{b_1}{c_0}f'(t) + \frac{b_0}{c_0}f(t)-\frac{c_2}{c_0}u''(t)-\frac{c_1}{c_0}u'(t) and g(t) = \frac{1}{1600}g''(t) + \frac{15}{1600}g'(t) -\frac{1}{1600}f(t)

And then putting their right hand sides equal each other, but this didn't really get me anywhere.

Just before posting this question I’ve been staring and trying for another 4 hours, and I feel I’m that my self-confidence is at an all-time low and I'm starting to ask myself if I really should be doing math at all (Yes, it's really a first world issue, I know).

REALLY grateful for any help!

 

[jackonelli]

Ok, sorry. Actually, putting the last two equations equal to each other really does give values to the coefficients. So I think that I managed to solve it after all This was only a sub-problem, so maybe I'll be back later...