<s>[INST] Exact Differential Equation - possible textbook mistake - $(3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0$ and $y(0)=2$ When solving the differential equation
$$
(3x^2+6xy^2) dx + (6x^2y+4y^2)dy = 0
$$
I verified that it is an Exact Equation. By saying that
$$
F_x = 3x^2+6xy^2\\
F_y = 6x^2y+4y^2
$$
and integrating both sides I found $$F(x,y) = 3x^2y^2 + \frac{4}{3}y^3+x^3 = C$$ and since $y(0)=2$ we have $$3x^2y^2+ \frac{4}{3}y^3+x^3 = \frac{32}{3}$$
The problem is that the textbook's solution is $$x^3+3x^2y^2+y^3=16$$
Where did I go wrong?
Thank you.
 [/INST] Your book solution is not correct $${ x }^{ 3 }+3{ x }^{ 2 }{ y }^{ 2 }+{ y }^{ 3 }=16\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+6{ x }^{ 2 }y{ y }^{ \prime  }+3{ y }^{ 2 }{ y }^{ \prime  }=0\\ 3{ x }^{ 2 }+6{ x }{ y }^{ 2 }+\frac { dy }{ dx } \left( 6{ x }^{ 2 }y+3{ y }^{ 2 } \right) =0\\ \left( 3{ x }^{ 2 }+6{ x }{ y }^{ 2 } \right) dx+\left( 6{ x }^{ 2 }y+\color{red}3{ y }^{ 2 } \right) dy=0\\ \\ \\ $$
which is a differ from original equation form($(3x^2+6xy^2) dx + (6x^2y+\color{red}4y^2)dy = 0$)
</s>