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 $A=0$.


My solution is the following, but I got stuck:








Then I made partial derivation:


$\frac{∂P}{∂y}: -(x^B+y^B)-y(By^{B-1})$

And then subtract:


But probably I did something wrong and I'm stuck and not sure where made I mistake.. Can you help me please?


2 Answers 2



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

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

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

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?

  • $\begingroup$ Can you please show me how you write it in this form? $\displaystyle \frac{dy}{dx} - \frac{y}{x} = \frac{(1+x)x^8}{(x^B+y^B)}$ $\endgroup$
    – Peter F.
    Oct 24, 2020 at 20:12
  • $\begingroup$ @PeterF. I edited. $\endgroup$
    – Math Lover
    Oct 24, 2020 at 20:19
  • $\begingroup$ Thank you, but I do not understand, how you take integrating factor.. $\endgroup$
    – Peter F.
    Oct 24, 2020 at 20:22
  • $\begingroup$ See if my answer a while ago helps math.stackexchange.com/questions/3879604/…. $\endgroup$
    – Math Lover
    Oct 24, 2020 at 20:24
  • $\begingroup$ Thanks it is more clear, but I thought that I shoult do integrating factor by following one of the principle on this site math24.net/using-integrating-factor ,,, I was just trying to do it in this way $\endgroup$
    – Peter F.
    Oct 24, 2020 at 20:28

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


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .