Differential Equations: The definite integral [integral sign from 0 -- infinity] te^[(4-s)t]dt represents the Laplace transform of which function?

Question

Differential Equations: The definite integral [integral sign from 0 --> infinity] te^[(4-s)t]dt represents the Laplace transform of which function?

anonymous · Answer

Equation in a better form: $$\int\limits_{0}^{\infty} te ^{(4-s)t}dt$$

anonymous · Answer

The integrand for the transform is e^(-st)*f(t). 
So we need to isolate the e^(-st) 
te^(4-s)t=te^(4t)*e^(-st) by the properties of exponents 
So we can see f(t) there as te^(4t)

anonymous · Answer

so that's it? whatever is left over after the e^(-st) was isolated is what was transformed?

anonymous · Answer

Yes

anonymous · Answer

awesome, thanks!