Evaluate the integral integral y e^-x\,dx.

We can use a substitution for this part. Let $u = -x$.
Then, differentiate $u$ with respect to $x$:

$$\frac{du}{dx} = -1$$

So, $du = -dx$, which means $dx = -du$.

Substitute $u$ and $dx$ into the integral:

$$\int e^{-x}\,dx = \int e^u (-du) = -\int e^u\,du$$

The integral of $e^u$ is $e^u$.

$$-\int e^u\,du = -e^u + C_1$$

Now, substitute back $u = -x$:

$$-e^{-x} + C_1$$

Multiply the result from Step 2 by $y$:

$$y(-e^{-x} + C_1) = -y e^{-x} + yC_1$$

Since $y$ is a constant, $yC_1$ is also an arbitrary constant, which we can denote as $C$.

$$-y e^{-x} + C$$