# What is a solution to the differential equation dy/dx=x/(x+y)?

Jan 31, 2017

See below.

#### Explanation:

Making $y = \lambda x$ and knowing that

$\mathrm{dy} = \lambda \mathrm{dx} + x d \lambda \to \frac{\mathrm{dy}}{\mathrm{dx}} = \lambda + x \frac{d \lambda}{\mathrm{dx}}$

we get at

$\lambda + x \frac{d \lambda}{\mathrm{dx}} = \frac{1}{\lambda + 1}$ or

$x \frac{d \lambda}{\mathrm{dx}} = - \frac{{\lambda}^{2} + \lambda - 1}{\lambda + 1}$ and now this differential equation is separable

$\frac{\mathrm{dx}}{x} = - \frac{\left(\lambda + 1\right) d \lambda}{{\lambda}^{2} + \lambda - 1}$

so

$\frac{\mathrm{dx}}{x} = - \frac{\left(\lambda + 1\right) d \lambda}{{\left(\lambda + \frac{1}{2}\right)}^{2} - \frac{5}{4}}$

giving

$\log x = - \frac{1}{10} \left(\left(\sqrt{5} + 5\right) \log \left(\sqrt{5} - 1 - 2 \lambda\right) - \left(\sqrt{5} - 5\right) \log \left(\sqrt{5} + 1 + 2 \lambda\right)\right) + C$

Now substituting,

$\log x = - \frac{1}{10} \left(\left(\sqrt{5} + 5\right) \log \left(\sqrt{5} - 1 - 2 \frac{y}{x}\right) - \left(\sqrt{5} - 5\right) \log \left(\sqrt{5} + 1 + 2 \frac{y}{x}\right)\right) + C$

so the solution appears in implicit form

$G \left(y \left(x\right) , x , C\right) = 0$