Error in solution to differential equation?

Question

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?

Answer 1

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 $$S' +2\frac Sw= \frac {2}{w^5}$$

Answer 2

$\frac{1}{2}w^5L^2=-\frac{w}{2}+Cw^3$

Multiply 2 on both sides and then divide $m^5$ on both sides gives:

$L^2=\frac{-w+Kw^3}{w^5}=\frac{-1+kw^2}{w^4}$

Answer 3

You will get the solution faster if you use $u(w)=\frac12w^2L^2$ as then $$ u'(w)=wL(w)^2+w^2L(w)L'(w)=\frac1{w^3} $$ which integrates directly to $$ u(w)=-\frac{1}{2w^2}+c\implies L(w)=\pm\frac1w\sqrt{C-\frac1{w^2}}=\pm\frac{\sqrt{Cw^2-1}}{w^2}. $$