Laplace Transform equation help

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SUMMARY

The discussion focuses on the Laplace Transform, specifically demonstrating that \(\mathcal{L}(f(ct)) = \frac{1}{c} F\left(\frac{s}{c}\right)\). The solution involves substituting \(u = ct\) into the integral definition of the Laplace Transform, leading to the conclusion that the transformed function scales appropriately with respect to the constant \(c\). The integration by parts method was considered but deemed unproductive for this specific problem.

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  • Basic concepts of function scaling in mathematical transformations
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[SOLVED] Laplace Transform

Homework Statement


Suppose F(s) = [tex]\displaystyle\mathcal{L}(f(t))[/tex]
Show that [tex]\displaystyle\mathcal{L}(f(ct)) = 1/c F(s/c)[/tex]

Homework Equations


The Attempt at a Solution



[tex]\displaystyle\mathcal{L}(f(t)) = \int_0^{inf} e^{-st} f(t) dt[/tex]
[tex]\displaystyle\mathcal{L}(f(ct)) = \int_0^{inf} e^{-st} f(ct) dt[/tex]
I'm not quite sure what to do after this...

I could play around with integration by parts, but in this case I don't think it yields anything useful
[tex]\frac{F(ct)}{c} e^{-st} - \int \frac{F(ct)}{c} (-s e^{-st} ) dt[/tex]

[tex]{F(t)e^{-st} + \int F(t) se^{-st} dt}{}[/tex]
 
Last edited:
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Suppose [tex]u=ct[/tex] then we get:

[tex]\int_0^{ \infty} e^{-st} f(ct)\ \mbox{d}t = \frac{1}{c} \int_0^{\infty}\ e^{\frac{-s}{c} u}\ f(u)\ \mbox{d}u = \frac{1}{c}\ F \left( \frac{s}{c} \right)[/tex]
 

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