Solve $y'=\frac{2}{x^2}-y^{2}$

Question

$$y'=\frac{2}{x^2}-y^{2}$$ $y=\frac{k}{x}$ is a solution, find the general solution

So we have riccati differential equation.

$y=z+\frac{k}{x}\iff z=y-\frac{k}{x}$

$y'=z'-\frac{k}{x^2}$

Plugin it to the ode we get:

$$z'-\frac{k}{x^2}=\frac{2}{x^2}-(z+\frac{k}{x})^2$$

$$z'=\frac{2-k^2+k}{x^2}-\frac{2kz}{x}-z^2$$

Which is again a riccati differential equation do I need to guess a particular $k$? for example $k=-1$?

Answer 1

Hint: if $y=\frac{k}{x}$ is a solution then pluging in the equation we get $$\\ -\frac { k }{ { x }^{ 2 } } =\frac { 2 }{ { x }^{ 2 } } -\frac { { k }^{ 2 } }{ { x }^{ 2 } } \\ { k }^{ 2 }-k-2=0\\ \left( k-2 \right) \left( k+1 \right) =0\\ $$