Differential equation with separation of the variables I have to slove the following differential equation with separation of the variables (without any initial conditions)
$xy'=\frac{1}{y^2}-y$
This is what I have done so far
$x\frac{dy}{dx}=\frac{1-y^3}{y^2}$
$\int \frac{dx}{x}=\int \frac{y^2 dy}{1-y^3}$
${\rm ln} \ \vert x \vert +c_1=-\frac{1}{3} {\rm ln} \ \vert 1-y^3 \vert +c_2 $
${\rm ln} \ \vert x \vert=-\frac{1}{3} {\rm ln} \ \vert 1-y^3 \vert +c \ $ with $c=c_2-c_1$
How shoud I go on? I don't know what to do with the absolute values. The solution should be $y(x)=(1-\frac{e^{3c}}{x^3})^{1/3}$. 
 A: The natural logarithm is one-to-one between the positive reals and the reals; exponentiate both members to eliminate the logarithms. The additive constant becomes a positive multiplicative constant.
As to the absolute values, you could start by ignoring them (so that your solution is technically only valid in the range where the argument of the absolute value is positive) and then check at the end whether your solution generalizes.
A: $$xy'(x)=\frac{1}{y(x)^2}-y(x)\Longleftrightarrow$$
$$x\cdot\frac{\text{d}y(x)}{\text{d}x}=\frac{1}{y(x)^2}-y(x)\Longleftrightarrow$$
$$\frac{\text{d}y(x)}{\text{d}x}=\frac{1-y(x)^3}{xy(x)^2}\Longleftrightarrow$$
$$\frac{\frac{\text{d}y(x)}{\text{d}x}\cdot y(x)^2}{1-y(x)^3}=\frac{1}{x}\Longleftrightarrow$$
$$\int\frac{\frac{\text{d}y(x)}{\text{d}x}\cdot y(x)^2}{1-y(x)^3}\space\text{d}x=\int\frac{1}{x}\space\text{d}x\Longleftrightarrow$$
$$-\frac{1}{3}\ln\left|-y(x)^3+1\right|=\ln|x|+\text{C}\Longleftrightarrow$$
$$\ln\left|-y(x)^3+1\right|=-3\ln|x|-3\text{C}\Longleftrightarrow$$
$$\ln\left|-y(x)^3+1\right|=\ln\left(\frac{1}{|x|^3}\right)-3\text{C}\Longleftrightarrow$$
$$\left|-y(x)^3+1\right|=e^{\ln\left(\frac{1}{|x|^3}\right)-3\text{C}}\Longleftrightarrow$$
$$\left|-y(x)^3+1\right|=\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow$$
$$1-y(x)^3=\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow\space\space\vee\space\space 1-y(x)^3=-\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow$$
$$-y(x)^3=\frac{e^{-3\text{C}}}{|x|^3}-1\Longleftrightarrow\space\space\vee\space\space -y(x)^3=-\frac{e^{-3\text{C}}}{|x|^3}-1\Longleftrightarrow$$
$$y(x)^3=1-\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow\space\space\vee\space\space y(x)^3=1+\frac{e^{-3\text{C}}}{|x|^3}\Longleftrightarrow$$
$$y(x)=\frac{\sqrt[3]{|x|^3-e^{-3\text{C}}}}{|x|}\space\space\vee\space\space y(x)=-\frac{\sqrt[3]{-1}\sqrt[3]{|x|^3+e^{-3\text{C}}}}{|x|}\space\space\vee\space\space y(x)=\frac{(-1)^{\frac{2}{3}}\sqrt[3]{|x|^3+e^{-3\text{C}}}}{|x|}$$
Where $\text{C}$ is an arbitrary constant
