- #1

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## Main Question or Discussion Point

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.