Help solving a Homogeneous Differential Equation? [Calculus 2]

Question

it's my first question here on Mathematics StackExchange, could anyone help me with solving the Homogeneous Differential Equation:

$$ \frac{ \mathrm d y}{ \mathrm d x}=-\frac{x^{2}-y^{2}}{x y} $$ I have already done: $$ \begin{array}{c} \mathrm d y=-\frac{x^{2}-y^{2}}{x y} \,\mathrm d x \\ x y \, \mathrm d y=-x^{2}-y^{2}\, \mathrm d x \end{array} $$ Let $y=v \cdot x$ $$ x\left(v \cdot x\right)\cdot \left(v\cdot \mathrm d x+x\cdot \mathrm d v\right)=-x^{2}-\left(v\cdot x\right)^{2} \, \mathrm d x $$ However, I am stuck there and don't know what to do next.

I do know that after I seperate the two variables, I take the integral, substitute $v = y/x$, and solve for $y$.

Thanks, much is appreciated!

Answer 1

since $\displaystyle\frac{dy}{dx}=-\frac{x^2-y^2}{xy}=\frac{y^2-x^2}{xy}=\frac{y}{x}-\frac{x}{y}$,

letting $y=vx$ gives $\displaystyle v+x\frac{dv}{dx}=v-\frac{1}{v}$ and so $\displaystyle x\frac{dv}{dx}=-\frac{1}{v}$.

Now separate the variables.

Answer 2

I suggest computing the inverted equation as $$ \dfrac{dx}{dy} = \dfrac{xy}{x^2-y^2}=\dfrac{x/y}{(x/y)^2-1} $$

with the change of variable $v = x/y$