<s>[INST] Solving a differential equation using an integrating factor 
I'm trying the solve the following equation: $ \left\{\begin{matrix}
(x^B+y^B)(xdy-ydx)=(1+x)x^9dx \\  y(-1)=A \end{matrix}\right. $
for $A=1$ and $A=0$.
$B\in2\mathbb{N}_0+1$

My solution is the following, but I got stuck:
$(x^B+y^B)(xdy-ydx)=(1+x)*x^9dx$
$x^B*xdy-y*x^Bdx+y^B*xdy-y*y^Bdx-(1+x)*x^9dx=0$
$x^B*xdy+y^B*xdy-y*x^Bdx-y*y^Bdx-(1+x)*x^9dx=0$
$xdy(x^B+y^B)+[-y(x^B+y^B)-(1+x)*x^9]dx=0$
where:
$Q(x,y)=x(x^B+y^B)dy$
$P(x,y)=[-y(x^B+y^B)-(1+x)*x^9]$
Then I made partial derivation:
$\frac{∂Q}{∂x}:(x^B+y^B)x(Bx^{B-1})$
$\frac{∂P}{∂y}: -(x^B+y^B)-y(By^{B-1})$
And then subtract:
$\frac{∂P}{∂y}-\frac{∂Q}{∂x}=-(x^B+y^B)-y(By^{B-1})-(x^B+y^B)-x(Bx^{B-1})$
But probably I did something wrong and
I'm stuck and not sure where made I mistake.. Can you help me please?
 [/INST] $(x^B+y^B)(xdy-ydx)=(1+x)x^9dx$
$xdy-ydx = \frac{(1+x)x^9}{(x^B+y^B)}dx$
$x \frac{dy}{dx} - y = \frac{(1+x)x^9}{(x^B+y^B)}$
$\displaystyle \frac{dy}{dx} - \frac{y}{x} = \frac{(1+x)x^8}{(x^B+y^B)}$
Based on the form of LHS, take integrating factor of $ \displaystyle \frac{1}{x}$.
$\displaystyle \frac {1}{x} \frac{dy}{dx}- \frac{y}{x^2} = \frac{(1+x)x^7}{(x^B+y^B)}$
$\displaystyle \frac{d}{dx} (\frac{y}{x}) = \frac{(1+x)x^{(7-B)}}{1+ (\frac{y}{x})^B}$
Substitute $u = \frac{y}{x}$
$\displaystyle ({1+ u^B}) du = ((1+x)x^{(7-B)}) dx$
Can you please take it from here?
</s>