Question

show that given equation is homogeneous and solve
(x^2+xy)dy=(x^2+y^2)dx

Answer 1

Equation $\frac{dy}{dx} = f(x,y)$ is homogeneous if the function $f(x,y)$ is homogeneous, that is

$f(tx,ty) = f(x,y)$ for any number $t$.

$$(x^2 + xy)dy = (x^2 + y^2)dx, \quad \frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy}$$

$f(tx,ty) = \frac{t^2x^2 + t^2y^2}{t^2x^2 + t^2xy} = \frac{x^2 + y^2}{x^2 + xy} = f(x,y)$ - equation is homogeneous.

Let $y = ux$, then $\frac{dy}{dx} = \frac{du}{dx} x + u$ and $\frac{dy}{dx} = \frac{x^2 + y^2}{x^2 + xy} = \frac{x^2 + u^2x^2}{x^2 + ux} = \frac{1 + u^2}{1 + u}$

$\frac{du}{dx} x + u = \frac{1 + u^2}{1 + u}, \quad \frac{du}{dx} x = \frac{1 + u^2}{1 + u} - u = \frac{1 - u}{1 + u}, \quad \frac{dx}{x} = \frac{1 + u}{1 - u} du,$

$$\frac{dx}{x} = \frac{1 - u + 2u}{1 - u} du, \quad \frac{dx}{x} = \left(1 - \frac{2(1 + \frac{1}{u - 1})}{1}\right) du$$ $$\int \frac{dx}{x} = \int \left(1 - \frac{2(1 + \frac{1}{u - 1})}{1}\right) du, \quad \ln(x) + c = -u - 2 \ln(u - 1),$$ $$u = -\ln(Ax) - 2 \ln(u - 1) = -\ln(Ax(u - 1)^2), \quad A = e^c = \text{const}$$ $$\frac{y}{x} = -\ln\left(Ax \left(\frac{y}{x} - 1\right)^2\right)$$