Differential equation help. Apparently I am doing something wrong because I did not get the answer in the book can someone spot my mistake? I will write out the problem and what I did below...

Question

anonymous · Answer

$$x\frac{ dy }{ dx }+y= \frac{ 1 }{ y^2} $$
Dividing by x across

$$\frac{ dy }{ dx }+\frac{y}{x}= \frac{ 1 }{x y^2} $$

Noticing its is a Bernoulli equation and letting N=-2

$$u=y^{3}$$

Then,

$$y=u^{\frac{1}{3}}$$

and
$$\frac{ dy }{ dx }= \frac{ 1 }{3}u^{\frac{-2}{3}}\ $$

Substituting back in
$$\frac{1}{3u^{\frac{2}{3}}}\frac{ du }{ dx }+\frac{u^\frac{1}{3}}{x}= \frac{1}{x\[(u^{\frac{ 1 }{ 3 }}) ^2$$}\     $$(u^{\frac{ 1 }{ 3 }}) \[(u^{\frac{ 1 }{ 3 }}) ^2$$\] (ignore the extra u^1/3 at the end i cant seem to get them to go away lol)

anonymous · Answer

multiplying through by 3u^(2/3) yeilds:

$$\frac{ du }{ dx }+\frac{ 3u}{ x }=\frac{ 3 }{ x }$$
finding the integrating factor
$$\mu(x)=\[e^{\int\limits_{}^{}\frac{ 1 }{ x }dx}$$ = x

which leaves 
$$\frac{ d }{ dx }3ux=3$$

solving for u

$$u(x)=\frac{ 3x+c }{ 3x } =1+\frac{ c }{ 3x }$$

anonymous · Answer

apparently the correct answer is 
$$1+\frac{ c}{ x^{-3} }$$

lgbasallote · Answer

integrating factor should be $$e^{\int \frac 3x dx}$$ right? i don't know if you did some trick afterwards though

anonymous · Answer

possibly that is my mistake... I'll have to go back and check

lgbasallote · Answer

yep. if you do that you'll arrive with $$1 + \frac c{x^3}$$

anonymous · Answer

ah thanks.. always something stupid like that I miss lol.