Hi thanks for your reply, but I am not sure what exactly do you mean by 
expand "then to factorize". 

I found that if I set the transfer function to be H = k / ( (s+p1) * (s+p2) 
), then the inverse laplace transform becomes:

k*e^(-p1*t) / (p1-p2) + k*e^(-p2*t) / (p1-p2)

*which is what I prefer.* It is weird to me that adding a constant 'k' 
change the form that Sympy chooses to show the result.

On Friday, January 1, 2016 at 2:46:19 AM UTC-5, Christophe Bal wrote:
>
> Hello. 
>
> Have you triez to expand 1nd then to factorize the formula ?
> Le 1 janv. 2016 03:42, "Ken" <[email protected] <javascript:>> a écrit :
>
>> I've just started learning Sympy. I wrote a few lines of code to perform 
>> a inverse laplace transform on a simple 2nd order transfunction:
>>
>> H(s) = 1 / ((s+p1) * (s+p2)).
>>
>> The result I got from Sympy is
>>
>> (e^(p1*t) - e^(p2*t))*e^-t*(p1+p2) / (p1 - p2)
>>
>> Is there a way to simplify this result to the one like in Maxima (e^-t*p1 
>> + e^-t*p2) / (p1-p2) ?
>>
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