Question

(8x-x^2y)dy+(x-xy^2)dx=0solve

Answer 1

Given that

$(8x-x^2y)dy+(x-xy^2)dx=0$

There is no specified method mentioned in the question. So, I used exact method to solve this.

Taking $x-xy$ as common

$(x-xy)((8-x)dy+(1-y)dx)=0$

$(8-x)dy+(1-y)dx=0$

$(1-y)dx+(8-x)dy=0$

The above equation is in the form of

$M(x,y)dx+N(x,y)dy=0$

$M(x,y)=(1-y)$ and $N(x,y)=(8-x)$

$\frac{\partial M}{\partial y}=-1$ and $\frac{\partial N}{\partial x}=-1$

$\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$

So, the equation is exact.

So, the general solution is

$\int Mdx+\int ($ terms of N not involving x)$dy=c$ (here c is the constant)

$\int (1-y)dx +\int 8dy=c$

$x-xy+8y=c$

So, the solution is $x-xy+8y=c$