Find the solution of the following Differential Equations by Exact (2y+xy)dx+2xdy=0,y(3)=2

Given:
${\displaystyle (2x+xy)dx+2xdy=0,y\left(3\right)=\sqrt{2}}$
this equation is form ${\displaystyle M(x,y)dx+N(x,y)dy=0}$
At this equation is exact if and only it
${\displaystyle \frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}}$
${\displaystyle M=2y+xy\frac{\partial M}{\partial y}=\frac{\partial }{\partial y}(2y+xy)=2+x}$
${\displaystyle N=2x\frac{\partial N}{\partial x}=2}$
Hence ${\displaystyle \frac{\partial M}{\partial y}\ne \frac{\partial N}{\partial x}}$
Now we find the integrating factor to made this equation exact.
${\displaystyle My-Nx=2+x-2=x}$
${\displaystyle \frac{My-Nx}{N}=\frac{x}{2x}=\frac{1}{2}}$
Hence ${\displaystyle I.F.=e^{\int \frac{1}{2}dx}=e^{\frac{x}{2}}}$
${\displaystyle \Rightarrow e^{\frac{x}{2}}(2y+xy)dx+2e^{\frac{x}{2}}xdy=0}$ (2)
$=2e^{2/2}y+x{e}^{x/2}y$
${\displaystyle N=2e^{\frac{x}{2}}x}$
${\displaystyle \frac{\partial M}{\partial y}=2e^{\frac{x}{2}}+x{e}^{\frac{x}{2}}}$
${\displaystyle \frac{\partial N}{\partial x}=2e^{\frac{x}{2}}y+2x\cdot e^{\frac{x}{2}}\cdot \frac{1}{2}=2e^{\frac{x}{2}}+x{e}^{\frac{x}{2}}}$
${\displaystyle \Rightarrow \frac{\partial M}{\partial y}=\frac{\partial M}{\partial x}}$
equation (2) is exact equation:
Now solution of equation
${\displaystyle \int Mdx+\int Mdy=c}$
${\displaystyle \int (2y{e}^{\frac{x}{2}}+xy{e}^{\frac{x}{2}})dx+\int dy=c}$