$x^2\frac{dy}{dx}+xy+1=0$

Question

Solve the following differential equation by the form of homogeneous equation. Letting $y=vx$

The equation: $x^2\frac{dy}{dx}+xy+1=0$
I can’t separate variables

My solution steps are: $x^2(v+x \frac{dv}{dx})+x^2v+1=0$ $(v+x \frac{dv}{dx})+v=\frac{-1}{x^2}$

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

How to separate variables x and v?

Answer 1

Instead, let $v = xy$ such that $\frac{dv}{dx} = y + x\frac{dy}{dx}$

$x^2 \frac{dy}{dx} + xy = x\frac{dv}{dx}$ and so $x^2\frac{dy}{dx} + xy + 1 = x\frac{dv}{dx} + 1 = 0$

You can solve $x\frac{dv}{dx} = -1$ by seperating.

Answer 2

Divide the proposed DE by $x$ in order to obtain \begin{align*} x^{2}y' + xy + 1 = 0 & \Longleftrightarrow xy' + y = -\frac{1}{x}\\\\ & \Longleftrightarrow (xy)' = -\frac{1}{x}\\\\ & \Longleftrightarrow xy = -\ln|x| + k\\\\ & \Longleftrightarrow y(x) = -\frac{\ln|x|}{x} + \frac{k}{x} \end{align*}

and we are done.

Hopefully this helps !