Answer
$y=\dfrac{1}{e^x}+\dfrac{c}{e^{2x}}$
Here, $c$ is an arbitrary constant.
Work Step by Step
Here, we have $e^x \dfrac{dy}{dx}+2e^x y=1$
or, $\dfrac{dy}{dx}+2 y=\dfrac{1}{e^x}$
The integrating factor is: $v(x)=e^{\int 2 dx}=e^x$
So, $(e^{2x}y)'=e^x$
In order to determine the general solution, isolate the x and y terms on one side and integrate both sides.
$\int (e^{2x}y)' dx=\int e^x dx$ ...(1)
Equation (1) gives:
$e^{2x}y=e^x+c$
Thus, the general solution is as follows:
$y=\dfrac{1}{e^x}+\dfrac{c}{e^{2x}}$
Here, $c$ is an arbitrary constant.