Question

What is the solution of the Homogeneous Differential Equation? :
$\frac{dy}{dx}=\frac{x^2+y^2-xy}{x^2}$ with $y\left(1\right)=0$

Answer 1

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

Explanation:

We have:

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

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

$y=vx$

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:

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

$\int \frac{1}{{(v-1)}^2}dv=\int \frac{1}{x}dx$

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

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

Using the initial condition, $-\frac{1}{v-1}=\ln \left|x\right|+C$, we get:

$-\frac{1}{0-1}=\ln \left|1\right|+C\Rightarrow 1$

Thus we have:

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

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

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

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

$=\frac{\ln \left|x\right|}{1+\ln \left|x\right|}$

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

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

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