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I'd like to get the inverse Laplace transform of a symbolic expression. The output of my code is equal to the input? How can I modify the code to obtain the correct result?

All k and V1 are positives.

InverseLaplaceTransform[(s + Subscript[k, 2] + Subscript[k, 12])/
 Subscript[V, 1][
  s^2 + s[Subscript[k, 1] + Subscript[k, 2] + Subscript[k, 12] + 
     Subscript[k, 21]] + Subscript[k, 1] Subscript[k, 2] + 
   Subscript[k, 1] Subscript[k, 12] + 
   Subscript[k, 2] Subscript[k, 21]], s, t]

enter image description here

Thank you so much in advance.

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1 Answer 1

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Don't use a square brackets.

$\mathcal{L}_s^{-1}\left[\frac{s+k_2+k_{12}}{V_1 \left(s^2+s \left(k_1+k_2+k_{12}+k_{21}\right)+k_1 k_2+k_1 k_{12}+k_2 k_{21}\right)}\right](t)$

InverseLaplaceTransform[
(s + Subscript[k, 2] + Subscript[k, 12])/(Subscript[V, 
1]*(s^2 + s*(Subscript[k, 1] + Subscript[k, 2] + Subscript[k, 12] + 
Subscript[k, 21]) + Subscript[k, 1] Subscript[k, 2] + 
Subscript[k, 1] Subscript[k, 12] + Subscript[k, 2] Subscript[k, 21])), s, t]

$\frac{e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}-\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_1-e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}+\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_1-e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}-\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_2+e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}+\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_2-e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}-\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_{12}+e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}+\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_{12}+e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}-\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_{21}-e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}+\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} k_{21}+e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}-\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}+e^{t \left(-\frac{k_1}{2}-\frac{k_2}{2}-\frac{k_{12}}{2}-\frac{k_{21}}{2}+\frac{1}{2} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}\right)} \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)}}{2 \sqrt{\left(k_1+k_2+k_{12}+k_{21}\right){}^2-4 \left(k_1 k_2+k_{21} k_2+k_1 k_{12}\right)} V_1}$

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  • $\begingroup$ very good! Thank you so much! $\endgroup$ Dec 7, 2017 at 14:04

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