What do you Know about ODE?

An exact ODE has the form: $A(x,y)dy+B(x,y)dx=0$ Where there is a $C^{2}$ function $F(x,y)$ such that $\frac{\partial F}{\partial y}=A(x,y)$ and $\frac{\partial F}{\partial x}=B(x,y)$ , which means that $\frac{\partial ^{2} F}{\partial x \partial y} = \frac{\partial A}{\partial x} = \frac{\partial B}{\partial y}$ . Being $F(x,y)=0$ the solution to the ODE as an implicit function. Using this, the ODE above can be written as: $(e^{2x}\cos(y)+x^2y)dy+(2e^{2x}\sin(y)+xy^2)dx=0$ And also we have $\frac{\partial}{\partial x}(e^{2x}\cos(y)+x^2y)=2e^{2x}\cos(y)+2xy=\frac{\partial}{\partial y}(2e^{2x}\sin(y)+xy^2)$ So the ODE is said to be exact. $\Rightarrow \frac{\partial F(x,y)}{\partial y}=e^{2x}\cos(y)+x^2y \wedge \frac{\partial F(x,y)}{\partial x} =2e^{2x}\sin(y)+xy^2$ With the first equation: $F(x,y)=e^{2x}\sin(y)+\frac{x^2y^2}{2}+C(x)$ Where $C(x)$ is an arbitrary function of $y$ . \ Replacing in the second equation: $2e^{2x}\sin(y)+xy^2+C'(x)=2e^{2x}+xy^2 \Rightarrow C'(x)=0 \Rightarrow C(x)=k$ $\Rightarrow F(x,y)=e^{2x}\sin(y)+\frac{x^2y^2}{2}+k$ So the solution of the ODE can be expressed as: $e^{2x}\sin(y)+\frac{x^2y^2}{2}+k=0$ And replacing the point $(0,\frac{\pi}{2})$ which belongs to the solution: $e^{2x}\sin(y)+\frac{x^2y^2}{2}=1$ So, replacing in the asked point $(x,\pi)$ : $e^{2x}\cdot 0+\frac{x^2\pi^2}{2}=1 \Rightarrow x=\frac{\sqrt{2}}{\pi}$