First order differential equation: did i solve this equation right So i'm trying to solve:

$$x^2\frac{dy}{dx} + 2xy = y^3$$
I'm given this differential equation, that Bernoulli equation:
  $$\frac{dy}{dx} + p(x)y = q(x)y^{n} $$

I think i've solved it and got 
$$ u = \frac{2}{5x} +Cx^4$$
I'm just not sure i am right i will show you how i get there but firstly...
This was part of another question which i've already solved

Show that if $y$ is the solution of the above Bernoulli differential
  equation and $u = y^{1−n}$, then $u$ satisfies the linear differential
  equation:
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$

Applying the chain rule to $u = y^{1-n}$ we obtain that 
\begin{align}
\frac{d u}{dx}(x)&= \frac{du}{dy}\cdot\frac{dy}{dx}\\
&= (1-n)y^{-n}\cdot\frac{dy}{dx}
\end{align}
Futhermore using the Bernoulli equation we have 
$$
\frac{dy}{dx}=q(x)y^n-p(x)y
$$
and 
\begin{align}
\frac{d u}{dx}&= (1-n)y^{-n}\cdot\frac{dy}{dx}\\
&=(1-n)y^{-n}\cdot q(x)y^n - (1-n)y^{-n}\cdot p(x) y\\
&=(1-n)q(x) -(1-n)p(x)y^{1-n}\\
&=(1-n)q(x) -(1-n)p(x)u
\end{align}
Hence U satisfies the equation
$$
\frac{du}{dx}+(1-n)p(x)u = (1-n)q(x)
$$

$$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that


*

*n = 3

*1- n = 1-3 = -2

*p(x) = $ \frac{2}{x}$

*q(x) = $x^{-2}$

*u = $y^{1-3} = y^{-2}$


Subbing these in...
$$      
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So... 
$$ \text{integrating  factor} = e^{\int p(x) \, dx} $$
 - p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating  factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6}  $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6}  $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6}  $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c  $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
 A: After correction, the solution to the initial differential equation is
$$y(x) = \pm \sqrt{ \frac{5x} {5Cx^5 + 2} }$$
Regarding the $u$ transformation: from the B.E. we know (1) $y' = q y^n - p y$. From the $u$ definition we know that (2) $u' = (1-n) y^{-n} y'$. Substitute $y'$ from (1) into (2) to recover the new ODE for $u$.
A: $$x^2\frac{dy}{dx} + 2xy = y^3$$
Divide both sides by $x^2$
$$\frac{dy}{dx} + \frac{2}{x}y = x^{-2} y^3$$
Consider
$$\frac{du}{dx} + (n-1)p(x)u = (1-n)q(x)$$
We know that


*

*n = 3

*1- n = 1-3 = -2

*p(x) = $ \frac{2}{x}$

*q(x) = $x^{-2}$

*u = $y^{1-3} = y^{-2}$


Subbing these in...
$$      
\frac{du}{dx} + (-2)\frac{2}{x}u = (-2)x^{-2}
$$
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
So... 
$$ \text{integrating  factor} = e^{\int p(x) \, dx} $$
 - p(x) dx = $-\frac{4}{x}$
$$ -4 \int \frac{1}{x} = -4log(x) = log (x^{-4}) $$
$$ \text{integrating  factor} = e^{log (x^{-4})}= x^{-4} = \frac{1}{x^4}$$
So multiply this to the equation
$$\frac{1}{x^4}\frac{du}{dx} + \left(\frac{-4}{x^5} \right)u = \frac{-2}{x^6}$$
So we want to solve
$$ \frac{d}{dx}\frac{1}{x^4}u = \frac{-2}{x^6}  $$
$$ \int \frac{d}{dx}\frac{1}{x^4}u = \int \frac{-2}{x^6}  $$
$$ \frac{1}{x^4}u = -2\int \frac{1}{x^6}  $$
$$ \frac{1}{x^4}u = -2\frac{1}{-5x^5} + c  $$
$$ \frac{1}{x^4}u = \frac{2}{5x^5} + c $$
$$ \therefore u= \frac{2}{5x} + cx^{4} $$
is this fine? Or do i need to somehow equate this y or sub $u=y^{1-n}$
As $$u=y^{-2}$$
$$\frac{1}{y^2}= \frac{2}{5x} + cx^{4} $$
$$y^2= \frac{5x}{2} + \frac{1}{cx^{4}} $$
$$y= \sqrt{\frac{5x}{2} + \frac{1}{cx^{4}}} $$
A: How's about...
$$
\frac{du}{dx} + \left(-\frac{4}{x}\right)u = (-2)x^{-2}
$$
$$
 \left(-\frac{4}{x}\right)u = (-2)x^{-1} -\frac{du}{dx}
$$
$$
 u = \frac{2}{4x} + \frac{x}{4}\frac{du}{dx}
$$
As $u=y^{1-n} = y^{-2}$
$$\frac{1}{y^2} =  \frac{2}{4x} + \frac{x}{4}\frac{du}{dx} $$
$$ y^2 = =  \frac{4x}{2} + \frac{4}{x}\frac{dx}{du} $$
$$ y  =  \sqrt{\frac{4x}{2} + \frac{4}{x}\frac{dx}{du}} $$
A: $$
x^2y' + 2xy = y^3
$$
first thing to notice is
$$
\dfrac{d}{dx}x^2y = x^2y' + 2xy
$$
so we have
$$
\dfrac{d}{dx}x^2y  = y^3
$$
let $v = x^2y$
we then have
$$
\dfrac{dv}{dx} = \left(\frac{v}{x^2}\right)^3 = \frac{1}{x^6}v^3
$$
