Is the following function even or odd? Let
$\frac{dx}{dw} = w^2x^4+1$
Show that $x(w)$ is an odd function.
While trying to solve the problem I came across the following theorem:

If $dy/dx=f(x,y)$ which is a function which is even then the solution $y_0$ is a odd function

However, before trying to prove the theorem, I am confused about the fact that whether $f(x,y)$ is even in x, y or both.
Can someone help me to see the intuition behind the problem?
 A: The intuition is that you want is to compare the differential equation "going to the right" from $x=0$, and the differential equation "going to the left" from $x=0$. That is, if $x=t$, increasing $t$ is "going to the right" and $\dfrac{dy}{dt}=f(t,y)$; if $x=-t$, increasing $t$ is "going to the left" and $\dfrac{dy}{dt}=-f(-t,y)$. Thus, if $f$ is even in $x$, the solution will "go down to the left" in the same way as it "goes up to the right", and hence be odd provided the initial condition is $y(0)=0$. If $f$ is odd in $x$, the solution will "go up to the left" in the same way as it "goes up to the right", and hence be even (regardless of the initial condition). You might want to make this a bit more rigorous.
A: Even/odd functions are decided by their explicit function graphs, viz cartesian $~ y=f(x)$  by seeing whether symmetry exists w.r.t. y-axis or origin respectively, but not by their differential equations.
When integrated with particular boundary conditions $(x,y)=(0,0)$ we should have
$f(x)= f(-x)$ for even functions and $f(-x)= -f(x)$ for odd functions
and their derivatives are respectively odd and even.. can also be checked through their series representations.
