Laplace transform of a composed function

Answered question



i'm here because i didn't found anything general on this topic, i need to know what is the expression of the laplace transform a composed function like that : L{f(x)g(x)}=L{f(g(x))}=??

my complete equation to transform is : x(t)=4×[a¨(t)cos(a(t))a˙(t)2sin(a(t))]+2500y(t)=4×[a¨(t)sin(a(t))a˙(t)2cos(a(t))]+3.7114×t2+2700

the goal is to isolate the a(t) function as a function of x(t) and y(x)

Thanks for your time,

Robin Luiz

Answer & Explanation



Expert2023-04-21Added 777 answers

To find the Laplace transform of a composite function like L{f(x)g(x)}, we can use the property that L{f(g(x))}=G(s)g'(s), where G(s) is the Laplace transform of f(g(x)).

Now, let's apply this property to the function x(t) given in the prompt: x(t)=4[-a''(t)sin(a(t))-a'(t)2cos(a(t))]+3.7774t2+2700

First, we need to take the Laplace transform of each term separately. The Laplace transform of 4[-a''(t)sin(a(t))] can be found using the property L{f''(t)}=s2F(s)-0-f'(0), and the Laplace transform of a'(t)2cos(a(t)) can be found using the property L{f'2(t)}=-F''(s).

So we have: L{4[-a''(t)sin(a(t))]}=4[s2L{sin(a(t))}+sa'(0)L{cos(a(t))}+a''(0)L{sin(a(t))}]L{a'(t)2cos(a(t))} 


Next, we need to take the Laplace transform of the remaining term, 3.7774t2+2700. The Laplace transform of tn is n!sn+1, so we have: L{3.7774t2+2700}=3.77742!s3+2700s

Putting it all together, we have: 



Note that we have used the initial conditions a'(0) and a''(0) to simplify some terms.

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