- #1
maistral
- 240
- 17
I'm inverting this:
Y = s2 + 15s + 17 / [(s+1)(s2 + 13s - 4)]
I'm using PF expansion,
A/(s+1) + Bs + C/(s2 + 13s - 4), I however keep on getting wrong answers, seeing how Runge-Kutta and Taylor approximation disagrees with my final equation.
My final equation is:
exp(-13/2t)[19/16cosh√(185)/2t + 13√(185)/74sinh√(185)/2t] - 3/16exp(-t) = y, and it's wrong (considering Runge-Kutta and Taylor approximation disagrees with it).
Obviously, something's wrong. What did I miss? I'm starting to think that the second term is.. well, there's something wrong with it (Bs + C term). I mean, the numerator is a quadratic, therefore it can't be that simple...NOTE: This is NOT homework. I did this to merely tickle my head. The original differential equation is y"+13y'-4y = 3exp(-t), y(0) = y'(0) = 1.
Y = s2 + 15s + 17 / [(s+1)(s2 + 13s - 4)]
I'm using PF expansion,
A/(s+1) + Bs + C/(s2 + 13s - 4), I however keep on getting wrong answers, seeing how Runge-Kutta and Taylor approximation disagrees with my final equation.
My final equation is:
exp(-13/2t)[19/16cosh√(185)/2t + 13√(185)/74sinh√(185)/2t] - 3/16exp(-t) = y, and it's wrong (considering Runge-Kutta and Taylor approximation disagrees with it).
Obviously, something's wrong. What did I miss? I'm starting to think that the second term is.. well, there's something wrong with it (Bs + C term). I mean, the numerator is a quadratic, therefore it can't be that simple...NOTE: This is NOT homework. I did this to merely tickle my head. The original differential equation is y"+13y'-4y = 3exp(-t), y(0) = y'(0) = 1.