What is the solution of the Homogeneous Differential Equation? : dy/dx=(x^2+y2^−xy)/(x^2) with y(1)=0

What is the solution of the Homogeneous Differential Equation? : $\frac{dy}{dx}=\frac{{x}^{2}+{y}^{2}-xy}{{x}^{2}}$ with y(1)=0
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Koen Henson
This is a First Order Nonlinear Ordinary Differential Equation. Let us attempt a substitution of the form:
$y=vx$
Differentiating wrt x and applying the product rule, we get:
$\frac{dy}{dx}=v+x\frac{dv}{dx}$
Substituting into the initial ODE we get:
$v+x\frac{dv}{dx}=\frac{{x}^{2}+{\left(vx\right)}^{2}-x\left(vx\right)}{{x}^{2}}$
Then assuming that $x\ne 0$ this simplifies to:
$v+x\frac{dv}{dx}=1+{v}^{2}-v$
$\therefore x\frac{dv}{dx}={v}^{2}-2v+1$
And we have reduced the initial ODE to a First Order Separable ODE, so we can collect terms and separate the variables to get:

Both integrals are standard, so we can integrate to get:
$-\frac{1}{v-1}=\mathrm{ln}|x|+C$
Using the initial condition, $y\left(1\right)=0⇒v\left(1\right)=0$, we get:
$-\frac{1}{0-1}=\mathrm{ln}|1|+C⇒1$
Thus we have:
$-\frac{1}{v-1}=\mathrm{ln}|x|+1$
$\therefore 1-v=\frac{1}{1+\mathrm{ln}|x|}$
$\therefore v=1-\frac{1}{1+\mathrm{ln}|x|}$

Then, we restore the substitution, to get the General Solution:
$\frac{y}{x}=\frac{\mathrm{ln}|x|}{1+\mathrm{ln}|x|}$
$\therefore y=\frac{x\mathrm{ln}|x|}{1+\mathrm{ln}|x|}$