Question #d88e8

Jan 19, 2017

$y \left(x\right) = a - \frac{c}{\left\mid x \right\mid}$

Explanation:

In the equation:

$x \frac{\mathrm{dy}}{\mathrm{dx}} + y = a$

we can separate the variables in this way:

$x \frac{\mathrm{dy}}{\mathrm{dx}} = a - y$

$\frac{\mathrm{dy}}{a - y} = \frac{\mathrm{dx}}{x}$

then integrate both sides:

$\int \frac{\mathrm{dy}}{a - y} = \int \frac{\mathrm{dx}}{x}$

$- \int \frac{d \left(a - y\right)}{a - y} = \int \frac{\mathrm{dx}}{x}$

$- \ln \left\mid a - y \right\mid = \ln \left\mid x \right\mid + C$

$\ln \left\mid a - y \right\mid = - \ln \left\mid x \right\mid + C$

We can now take the exponential of both sides posing $c = {e}^{C} > 0$:

$\left\mid a - y \right\mid = {e}^{- \ln \left\mid x \right\mid + C} = {e}^{C} / {e}^{\ln \left\mid x \right\mid} = \frac{c}{\left\mid x \right\mid}$

If we make the hypothesis $y < a$ we have:

$a - y = \frac{c}{\left\mid x \right\mid}$

$y = a - \frac{c}{\left\mid x \right\mid}$

which is in fact always less than $a$