Error in solution to differential equation? I'm trying to solve $w^5L\frac{dL}{dw} = 1 - w^4L^2$. I attempted the substitution $M = \frac12w^5L^2$, so that $M' = \frac52w^4L^2 + w^5L'L$. Then
$$
w^5L'L = M' - \frac52 w^4L^2 = M' - \frac{5M}w \\
1 - w^4L^2 = 1 - \frac{2M}w
$$
so the equation becomes
$$
M' = 1 + \frac{3M}w
$$
which can be solved to give
$$
M = c_1w^3 - \frac12w
$$
and so
$$
w^5L^2 = c_1w^2(2w - 1) \\
L = \pm\sqrt{\frac{c_1(2w-1)}{w^3}}
$$
however, apparently, I'm supposed to get
$$
L = \pm\frac{\sqrt{c_1w^2 - 1}}{w^2}
$$
so it seems I'm missing a factor of $w$ somewhere, or something, but I'm not quite sure. Where am I going wrong?

I checked everything and everything was ok except here 
$$M = c_1w^3 - \frac12w$$
$$w^5L^2 = 2c_1w^3 -w$$
$$w^2L^2 = 2c_1 -\frac 1 {w^2}$$
$$\pm Lw= \sqrt{2c_1 -\frac 1 {w^2}}$$
$$ L= \pm \frac 1 {w^2}\sqrt{2c_1w^2 -1}$$
$$ \boxed{L= \pm \frac  {\sqrt{cw^2 -1}}{w^2}}$$
It's more simple to substitute $S=L^2$
$$w^5L\frac{dL}{dw} = 1 - w^4L^2$$
$$w^5LL' = 1 - w^4L^2$$
$$w^5\frac 12 (L^2)' = 1 - w^4L^2$$
$$w^5\frac 12 S' = 1 - w^4S$$
$$S' = \frac {2}{w^5}(1 - w^4S)$$
It's a linear first ode
$$S' +2\frac Sw= \frac {2}{w^5}$$
