# Identify the following differential equation and hence solve it y'=−4/x^2−y/x+y^2 ?

Identify the following differential equation and hence solve it $y\prime =-\frac{4}{{x}^{2}}-\frac{y}{x}+{y}^{2}$?
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Nolan Tyler
We have:
$y\prime =-\frac{4}{{x}^{2}}-\frac{y}{x}+{y}^{2}$.... [A]
This is a non-linear first order Differential Equation. We can attempt a substitution:
$v=xy⇔y=\frac{v}{x}$
And differentiating using the product rule we get:

And substituting into the DE [A] we get:
$\frac{\frac{dv}{dx}-y}{x}=-\frac{4}{{x}^{2}}-\frac{\frac{v}{x}}{x}+{\left(\frac{v}{x}\right)}^{2}$
$\therefore \frac{dv}{dx}-\frac{v}{x}=-\frac{4}{x}-\frac{v}{x}+\frac{{v}^{2}}{x}$
$\therefore \frac{dv}{dx}=\frac{{v}^{2}-4}{x}$
Which is now separable, so we can collect terms, and "separate the variables" to get:

And we can integrate to get:
$\frac{1}{2}{\mathrm{tanh}}^{-1}\left(\frac{v}{2}\right)=\mathrm{ln}x+C$
$\therefore {\mathrm{tanh}}^{-1}\left(\frac{v}{2}\right)=2\mathrm{ln}x+2C$
$\therefore \frac{v}{2}=\mathrm{tanh}\left(2\mathrm{ln}x+A\right)$
$\therefore v=2\mathrm{tanh}\left(2\mathrm{ln}x+A\right)$
And restoring the substitution:
$xy=2\mathrm{tanh}\left(2\mathrm{ln}x+A\right)$
$y=\frac{2}{x}\mathrm{tanh}\left(2\mathrm{ln}x+A\right)$