Solving a differential equation using an integrating factor

Question

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?

Answer 1

$(x^B+y^B)(xdy-ydx)=(1+x)x^9dx$

$xdy-ydx = \frac{(1+x)x^9}{(x^B+y^B)}dx$

$x \frac{dy}{dx} - y = \frac{(1+x)x^9}{(x^B+y^B)}$

$\displaystyle \frac{dy}{dx} - \frac{y}{x} = \frac{(1+x)x^8}{(x^B+y^B)}$

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?

Answer 2

$$(x^B+y^B)(xdy-ydx)=(1+x)x^9dx $$ Duivide by $x^2$: $$(x^B+y^B)d\dfrac yx=(1+x)x^7dx $$ Divode by $x^B$: $$\left(1+\left(\dfrac y x\right)^B \right)d\dfrac yx=\dfrac {(1+x)x^7}{x^B}dx $$ Integrate.