Convolution Integral and Differential Equation

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To solve the convolution integral and differential equation problem, start by applying the convolution theorem, which states that the Laplace transform of the convolution of two functions equals the product of their individual Laplace transforms. In this case, the functions involved are y(t-w) and e^{-10w}. The integral provided represents the convolution of y(t) and e^{-10t}. To find Y(s), take the Laplace transform of the entire differential equation, leading to an algebraic equation in Y(s) that can be solved for Y(s) in terms of s. Finally, apply the inverse Laplace transform to obtain y(t).
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I am really confused where to start with this problem. I know about convolutions somewhat. We have done them a little. Where is a good place to begin with this problem?
 
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What does the convolution theorem state about two functions? What are those two functions in this problem?

BiP
 
if f and g are piecewise continuous then the convolution of the two is
\int f(\tau)g(t-\tau)d\tau from 0 to t
or
\int f(t-\tau)g(\tau)d\tau from 0 to t

and

F(s)G(s)=Laplace Transform of the convolution of f and g


in this case y(t-w) is f(t-\tau) and e^{-10w} is g(\tau)

So now how can I use this too get y(t)?
 
Do you see that the integral in the problem you gave is the convolution of y(t) and e^{-10t ?

How can you apply the convolution theorem? You will obtain an algebraic equation in s. Solve it for Y(s).

BiP
 
So what I've figured so far is s*Y(s)-25* inverse laplace of {Y(s)G(s)} = 1

How can I solve for Y(s) when it is in the inverse laplace transform?
 
Taking the Laplace transform of every term in the differential equation should give you an algebraic equation in Y(s). Solve for this in terms of s, then take the inverse transform.

BiP
 

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