How Can You Find the Inverse Laplace Transform of \(\frac{s}{{(s + 4)^4}}\)?

[Benny]

Hi I'm having problems find the inverse Laplace transform of $$\frac{s}{{\left( {s + 4} \right)^4 }}$$ via a table/look up method.

In another question part I found the inverse Laplace transform of (s+4)^-4 by considering $$\frac{{d^3 }}{{db^3 }}\left[ {\left( {s + b} \right)^{ - 1} } \right]$$ so that the problem essentially reduced to finding the inverse Laplace transform of 1/(s+b). After that I just set b = 4 to get the answer.

But I can't think of a way to do this one (the one in the first sentence of this post). I have $$L^{ - 1} \left\{ {\frac{1}{{\left( {s + 4} \right)^4 }};s \to t} \right\} = \frac{{t^3 }}{6}e^{ - 2t}$$ but I don't see a way to use it to find the inverse Laplace transform of s(s+4)^-4. Can someone please help me out? Thanks.

 

[mighty2000]

Hi there,

I can definitely help you with finding the inverse Laplace transform of \frac{s}{{\left( {s + 4} \right)^4 }}. First, let's rewrite the expression as \frac{s}{(s+4)^3}\cdot \frac{1}{s+4}. Now, we can use the property of the Laplace transform that states L\{tf(t)\} = -F'(s), where F(s) is the Laplace transform of f(t). In this case, we have f(t) = \frac{1}{s+4} and F(s) = e^{-4s}. Therefore, by applying the property, we get L\{t\cdot \frac{1}{s+4}\} = -\frac{d}{ds}(e^{-4s}) = 4e^{-4s}.

Now, we are left with finding the inverse Laplace transform of \frac{s}{(s+4)^3}\cdot 4e^{-4s}. We can use another property of the Laplace transform that states L\{t^n\} = \frac{n!}{s^{n+1}}. In this case, we have n=2 and we get L\{t^2\} = \frac{2!}{s^{2+1}} = \frac{2}{s^3}. Therefore, by applying this property, we get L\{\frac{s}{(s+4)^3}\} = -\frac{d}{ds}(\frac{2}{s^3}) = \frac{6}{s^4}.

Finally, we can combine the two results to get the inverse Laplace transform of \frac{s}{{\left( {s + 4} \right)^4 }} as L^{-1}\{\frac{s}{(s+4)^4}\} = L^{-1}\{\frac{s}{(s+4)^3}\cdot \frac{1}{s+4}\} = L^{-1}\{t\cdot \frac{1}{s+4}\}\cdot L^{-1}\{\frac{s}{(s+4)^3}\} = 4e^{-4t}\cdot \frac{6}{t^3} = \frac{24}{t^3}e^{-4t}.

I hope this helps you with finding the inverse