Solution to a differential equation involving inseparable variables.

Question

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

Answer 1

$$\frac{dy}{dx} - \epsilon y = x$$

$$\frac{dy}{dx} = x + \epsilon y$$

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 $$y = (1 + \frac{1}{\epsilon^2 })e^{\epsilon x} - \frac{x}{\epsilon} - \frac{1}{\epsilon^2}$$