Find an explicit solution for $xy'+(x+\frac {1}{3})y=e^{-2x}y^{-2}$

Question

Beginning with the differential equation: $$xy'+(x+\frac {1}{3})y=e^{-2x}y^{-2}$$ Divide by $x$: $$y'+y+\frac {y}{3x}=e^{-2x}y^{-2}x^{-1}$$

Dividing by $y^{-2}$ gives: $$y^2y'+y^2y+\frac {1}{3xy}=e^{-2x}x^{-1}$$ Normally, for Bernoulli equations, one would substitute such that $v=y^2$ and $v'=2yy'$ but I seem to have simplified poorly.

Answer 1

$$xy'+(x+\frac {1}{3})y=e^{-2x}y^{-2}$$ Multiply by $3y^2$: $$3xy^2y'+(3x+1)y^3=3e^{-2x}$$ Let $\quad Y=y^3$ : $$xY'+(3x+1)Y=3e^{-2x}$$ This is a linear ODE easy to solve : $$Y=\frac{c}{xe^{3x}}+\frac{3}{xe^{2x}}$$ $$y(x)=\left(\frac{c}{xe^{3x}}+\frac{3}{xe^{2x}} \right)^{1/3}$$

Answer 2

The ODE given is

$$xy'+ \left(x+\dfrac 13\right)y=e^{-2x}y^{-2}$$

I shall also divide it by $x$. So, the ODE becomes:

$$y'+\left(1+\dfrac{1}{3x}\right)y =\dfrac{e^{-2x}}{x}y^{-2}\tag 1$$

This is Bernoulli's ODE.

We substitute $v=y^{1-(-2)}=y^3$.

So, $\dfrac{\mathrm dy}{\mathrm dx}=\dfrac{1}{3v^{(2/3)}}\dfrac{\mathrm dv}{\mathrm dx}$

$$\begin{align}(1)&\implies \dfrac{1}{3v^{(2/3)}}\dfrac{\mathrm dv}{\mathrm dx}+\left(1+\dfrac{1}{3x}\right)v^{1/3}=\dfrac{e^{-2x}}{x}v^{-2/3}\\ &\implies \dfrac{\mathrm dv}{\mathrm dx}+ 3\left(1+\dfrac{1}{3x}\right)v=3\dfrac{e^{-2x}}{x}\end{align}$$

I think you can take it from here.