Answer
$I(x,y)=x^3y^{-2}$
Work Step by Step
We are given
$y(5xy^2+4)dx+x(xy^2-1)dy=0$
Multiplying equation (1) by $I(x)=x^ry^s$
$x^ry^s(y(5xy^2+4)dx+x(xy^2-1)dy)=0$
$x^ry^sy(5xy^2+4)dx+x^ry^sx(xy^2-1)dy=0$
Here, $M(x,y)=x^ry^sy(5xy^2+4)$
$N(x,y)=x^ry^sx(xy^2-1)$
$\frac{\partial M}{\partial y}=x^ry^s((s(5xy^2+4)x^2+15xy^2+4)$
$\frac{\partial N}{\partial x}=x^{r}y^s(r(xy^2-1)+2xy^2-1)$
The given diffirential equation is exact
So $\frac{\partial N}{\partial x}=\frac{\partial M}{\partial y}$
$x^ry^s((s(5xy^2+4)x^2+15xy^2+4)=x^{r}y^s(r(xy^2-1)+2xy^2-1)$
$(5s+15)xy^2+4s+4=(r+2)xy^2-r-1$
$5(s+3)=r+2$
$4(s+1)=-(r+1)$
$\rightarrow r=3, s=-2$
Hence here,
$I(x,y)=x^3y^{-2}$
