What Is the Inverse Laplace Transform of Y(s)/(s+1)?

rforrevenge
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Homework Statement



What is the inverse Laplace transform of Y(s)/(s+1) ?

Homework Equations





The Attempt at a Solution



I think it is e^(-t) * y(t). Am i right?
 
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If * denotes convolution then you're right.
 
I'm posting your PM here because it's against PF rules to address homework (or homework-type) questions by PM:

rforrevenge said:
First of all thank you for taking the time to answer my question.The Y(s)/s+1 is a part of a Laplace equation and i have to find out which is the DE that gave that Laplace equation.So i inverse-Laplace the given equation to find the DE. My answer is still correct right?

Thank you and sorry for bothering you,
Rforrevenge

P.S:Sorry for my bad English

With regards to the above, I don't think I can answer it definitively because I don't know what is the original question, so clearly I can't tell if you are right.
 
the original question is that i have to find the DE of that Laplace transform
s*Y(s)+Y(s)/s+1=X(s)
 

Thread 'Prove that the integral is equal to ##\pi^2/8##'

Prove $$\int\limits_0^{\sqrt2/4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx = \frac{\pi^2}{8}.$$ Let $$I = \int\limits_0^{\sqrt 2 / 4}\frac{1}{\sqrt{x-x^2}}\arcsin\sqrt{\frac{(x-1)\left(x-1+x\sqrt{9-16x}\right)}{1-2x}} \, \mathrm dx. \tag{1}$$ The representation integral of ##\arcsin## is $$\arcsin u = \int\limits_{0}^{1} \frac{\mathrm dt}{\sqrt{1-t^2}}, \qquad 0 \leqslant u \leqslant 1.$$ Plugging identity above into ##(1)## with ##u...
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