Does anyone know how to find a general solution (or a particular solution, if a general solution is not feasible) to the following differential equation?
$$y'=\frac{y^2}{x}-1$$
This isn't separable, and Laplace transforms don't work out nicely. I've also tried a whole bunch of substitutions, none of which have led to a more manageable differential equation. Any ideas?
On
You can try to solve a second order linear equation instead
Except that it has variable coefficient but maybe you can work it..
$$y'=\frac{y^2}{x}-1$$ Substitute $v=\frac yx$
The equation becomes $$v'=v^2-\frac 1x-\frac vx$$ Now substitute $v=-\frac {u'}u$ $$u''x+u'-u=0$$ Maybe you can try now the Lpalce transform or a serie solution ?
It's not entirely clear to me why this works, or if there's any possible way of generalising it, but if you put (bear with me) $$ y= -x\frac{u'}{u}, $$ you find that the equation becomes $$ \frac{u-u'-xu''}{u} = 0, $$ which is a nice ordinary second-order ODE, related to Bessel's equation. It has solutions $$ u = A I_0(2\sqrt{x})+ BK_0(2\sqrt{x}), $$ from which it is easy to get $y$ when you know that $I_0'(z) = I_1(z) $, $ K_0'(z) = -K_1(z) $.