# What is the solution of the Homogeneous Differential Equation? : dy/dx = (x^2+y^2-xy)/x^2 with y(1)=0

May 17, 2018

$y = \frac{x \ln | x |}{1 + \ln | x |}$

#### Explanation:

We have:

$\frac{\mathrm{dy}}{\mathrm{dx}} = \frac{{x}^{2} + {y}^{2} - x y}{x} ^ 2$ with $y \left(1\right) = 0$

Which is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:

$y = v x$

Differentiating wrt $x$ and applying the product rule, we get:

$\frac{\mathrm{dy}}{\mathrm{dx}} = v + x \frac{\mathrm{dv}}{\mathrm{dx}}$

Substituting into the initial ODE we get:

$v + x \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{{x}^{2} + {\left(v x\right)}^{2} - x \left(v x\right)}{x} ^ 2$

Then assuming that $x \ne 0$ this simplifies to:

$v + x \frac{\mathrm{dv}}{\mathrm{dx}} = 1 + {v}^{2} - v$

$\therefore x \frac{\mathrm{dv}}{\mathrm{dx}} = {v}^{2} - 2 v + 1$

And we have reduced the initial ODE to a First Order Separable ODE, so we can collect terms and separate the variables to get:

$\int \setminus \frac{1}{{v}^{2} - 2 v + 1} \setminus \mathrm{dv} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$

$\int \setminus \frac{1}{v - 1} ^ 2 \setminus \mathrm{dv} = \int \setminus \frac{1}{x} \setminus \mathrm{dx}$

Both integrals are standard, so we can integrate to get:

$- \frac{1}{v - 1} = \ln | x | + C$

Using the initial condition, $y \left(1\right) = 0 \implies v \left(1\right) = 0$, we get:

$- \frac{1}{0 - 1} = \ln | 1 | + C \implies 1$

Thus we have:

$- \frac{1}{v - 1} = \ln | x | + 1$

$\therefore 1 - v = \frac{1}{1 + \ln | x |}$

$\therefore v = 1 - \frac{1}{1 + \ln | x |}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{1 + \ln | x | - 1}{1 + \ln | x |}$

$\setminus \setminus \setminus \setminus \setminus \setminus \setminus = \frac{\ln | x |}{1 + \ln | x |}$

Then, we restore the substitution, to get the General Solution:

$\frac{y}{x} = \frac{\ln | x |}{1 + \ln | x |}$

$\therefore y = \frac{x \ln | x |}{1 + \ln | x |}$