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Solve $$y(x)=e^x(1+\int_{0}^{x}e^{-t}y(t) dt)$$
Here's what I did: take derivative on both sides, $y'(x)=e^x+e^x \int_{0}^{x}e^{-t}y(t)dt +e^xe^{-x}y(x) \Rightarrow y'(x)=e^x+y(x)-e^x+y(x) \Rightarrow y'=2y$. Hence the solution looks like $y(x)=e^{2x}+C$. But I wonder how to solve this equation using Laplace transform and convolution. Any idea?
Your equation can be written as: $$ y(x) = e^{x} + \int_{0}^{x} e^{x-t}y(t)dt = \exp(x) + [\exp*y](x). $$ Then, if $ Y(s) $ is the Laplace transform of $ y(x) $, you get $$ Y(s) = \dfrac{1}{s-1} + \dfrac{ Y(s) }{s-1} \implies Y(s)=\dfrac{1}{s-2} $$ which means $ y(x) = e^{2x} $.
