Solution to a differential equation involving inseparable variables. What is the solution for the following DE 
$\frac{dy}{dx} - \epsilon{y} = x$ 
Where $0 \leq x \leq 1$ and  initial condition y = 1 when x = 0 and where $\epsilon$ is any positive parameter
 A: $$\frac{dy}{dx} - \epsilon y = x$$
$$\frac{dy}{dx}  = x + \epsilon y$$
Let $v = x + \epsilon y$
$$\frac{dy}{dx}  = v$$
Differentiating $v = x + \epsilon y$
$$\frac{dv}{dx} = 1 +\epsilon\frac{dy}{dx}$$
$$\frac{dy}{dx} = \frac{\frac{dv}{dx} - 1}{\epsilon}$$
Substituting $\frac{dy}{dx}$ we have $$  \frac{\frac{dv}{dx} - 1}{\epsilon} = v$$
$$  \frac{dv}{dx} = \epsilon v + 1$$ 
Dividing by $\epsilon v + 1$ and multiplying by $dx$
$$\frac{1}{\epsilon v + 1} {dv} = dx$$
Integrating both sides
$$\int\frac{1}{\epsilon v + 1} {dv} = \int dx$$
$$\frac{1}{\epsilon}ln(\epsilon v + 1) = x + C $$
$$ln(\epsilon v + 1) = \epsilon (x + C) $$
$$ln(\epsilon v + 1) = \epsilon x + \epsilon C $$
$\epsilon C$ is just another constant (C)
$$ln(\epsilon v + 1) = \epsilon x + C $$
Raising both sides to the power $e$
$$e^{ln(\epsilon v + 1)} = e^{\epsilon x + C}$$
$$e^{ln(\epsilon v + 1)} = e^{\epsilon x}  e^C$$
$e^C$ is just another constant $C$.
$$e^{ln(\epsilon v + 1)} = Ce^{\epsilon x} $$
$$\epsilon v + 1 = Ce^{\epsilon x} $$
Substiuting $v = x + \epsilon y$
$$\epsilon(x + \epsilon y) + 1 = Ce^{\epsilon x}$$
$$\epsilon x + \epsilon^2 y + 1 = Ce^{\epsilon x} $$
$$\epsilon x + \epsilon^2 y = Ce^{\epsilon x} - 1$$
$$\epsilon^2 y = Ce^{\epsilon x} - {\epsilon x}  - 1$$
Generic solution
$$y = C\frac{e^{\epsilon x}}{\epsilon^2 } - \frac{x}{\epsilon} - \frac{1}{\epsilon^2} $$
Using the initial condition $y= 1$ when $x = 0$ we calculate for $C$
$$C\frac{e^0}{\epsilon^2 } - \frac{0}{\epsilon} - \frac{1}{\epsilon^2} = 1$$
$$\frac{C - 1}{\epsilon^2 }  = 1$$
$$C= 1 + {\epsilon^2 }$$
By substituting C the exact solution is
   $$y = (1 + \frac{1}{\epsilon^2 })e^{\epsilon x} - \frac{x}{\epsilon} - \frac{1}{\epsilon^2}$$
