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