1
$\begingroup$

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?

$\endgroup$
1
$\begingroup$

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} $.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.