We are given the differential equation $y' = \frac{x^2 + 2y^2}{xy}$ and we are asked to solve it. This equation appears to be homogeneous.

1. Problem Description

We are given the differential equation $y' = \frac{x^2 + 2y^2}{xy}$ and we are asked to solve it. This equation appears to be homogeneous.

2. Solution Steps

Let $y = vx$, where $v$ is a function of $x$. Then, $\frac{dy}{dx} = v + x\frac{dv}{dx}$.

The given differential equation is:

$y' = \frac{x^2 + 2y^2}{xy}$

Substituting $y = vx$ and $y' = v + x\frac{dv}{dx}$ into the equation, we have:

$v + x\frac{dv}{dx} = \frac{x^2 + 2(vx)^2}{x(vx)} = \frac{x^2 + 2v^2x^2}{vx^2} = \frac{x^2(1+2v^2)}{vx^2} = \frac{1+2v^2}{v} = \frac{1}{v} + 2v$

Thus,

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

Separating variables, we get:

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

Integrating both sides, we have:

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

Let $u = 1+v^2$, so $du = 2v dv$. Then $\frac{1}{2}du = v dv$.

$\int \frac{1}{2u} du = \int \frac{1}{x} dx$

$\frac{1}{2}\ln|u| = \ln|x| + C$

$\frac{1}{2}\ln(1+v^2) = \ln|x| + C$

$\ln(1+v^2) = 2\ln|x| + 2C$

$\ln(1+v^2) = \ln(x^2) + \ln(e^{2C})$

$\ln(1+v^2) = \ln(x^2e^{2C})$

$1+v^2 = x^2e^{2C}$

$1+v^2 = Kx^2$, where $K = e^{2C}$ is a constant.

Substituting $v = \frac{y}{x}$, we have:

$1 + (\frac{y}{x})^2 = Kx^2$

$1 + \frac{y^2}{x^2} = Kx^2$

Multiplying by $x^2$, we get:

$x^2 + y^2 = Kx^4$

$y^2 = Kx^4 - x^2$

$y^2 = x^2(Kx^2 - 1)$

$y = \pm x\sqrt{Kx^2 - 1}$

3. Final Answer

$y^2 = x^2(Kx^2 - 1)$, where $K$ is a constant. Or $y = \pm x \sqrt{Kx^2 - 1}$, where $K$ is a constant.