When to solve for x instead of y when solving ODE The question asked to solve the ode:
$ydx-4(x+y^6)dy=0$
The answer given is $x=2y^6+cy^4$ isn't that cheating to solve for x instead of y? The section of the textbook this appears in, is linear equations. Obviously the $y^6$ messes things up, but maybe you can use algebra or a different solving technique all together. How do you know to solve for x instead of y? 
 A: $$y\frac{dx}{dy}-4(x+y^6)=0$$
Change of notations : $y=X$ and $x=Y$
$$X\frac{dY}{dX}-4(Y+X^6)=0$$
$$\frac{dY}{dX}-4\frac{Y}{X}=4X^5$$
This is a linear ODE, which solving leads to :
$$Y=2X^6+cX^4$$
$$x=2y^6+cy^4$$
If you want $y(x)$, solve for $t=y^2$ the third degree equation $$2t^3+ct^2-x=0$$
A: You just remove differentials "${\rm d}x$" and "${\rm d}y$". If it is expressible in explicit form try it otherwise leave it in implicit form.
A: Professors in my country also solve ODEs for $x$ instead of $y$ if it is easier to do it that way, for example ODE:
\begin{align}y'(x+y^2 )= y^2 & \text{ (1)}\end{align}
Eq(1) is not linear regarding x as the independent variable, but it is linear regarding x as a function of 'independent' variable y. Rearrange Eq(1) gives:
\begin{align}(x+y^2)dy = ydx ,\end{align}
Before dividing both sides by $dy$ to yield $dx/dy = x'$, we have to check whether $dy ≠ 0 → y ≠ C$, C is constant, is a solution of Eq(1). In this case $y = 0$ is indeed a solution of Eq(1). So, with $y ≠ 0$, divide both sides by $dy$:
\begin{align}x + y^2 = y \frac{dx}{dy} ↔ x + y^2 = yx'\end{align}
Next, divide both sides by y, which is nonzero from above, gives:
\begin{align}x' - \frac{x}{y} = y^2 & \text{  (2), which is linear.} \end{align}
The integrating factor: I = $e^{∫\frac{-1}{y} dy}$ = $e^{-ln\lvert y\rvert}$ = $\lvert y\rvert^{-1}$. Multiply both side Eq(2) by I, gives:
\begin{align}\frac{dx}{dy}[\lvert y\rvert^{-1}x] = \lvert y\rvert^{-1} y ↔ x = \lvert y\rvert∫\frac{y}{|y|}  dy\end{align}
There is an argument here, you can eliminate the absolute signs because |y| outside and under the integral symbol always have the same sign:
\begin{align}↔ x = y∫dy ↔ x = y(y + C_1)\end{align}
In conclusion:
\begin{align}Eq(1) ↔  [ \begin{array}
y = 0 \\ x = y(y + C_1) \end{array}\ \end{align} (I do not know how to code the big left square braket, the program does not give y =0, sorry for inconvenience).
** However, I find this method a little ambiguous. When we swap the role of $y$ for $x$($x$ as a function of $y$), do we automatically assume there exists an inverse function $y = f^{-1}(x)$ ? If the answer is Yes then how do we know that this inverse function really exists? Not any function has an inverse function. To have an inverse function $y = f^{-1}(x),  y = f(x)$ has to be a one-to-one function(see Section 1.6, Calculus Early Transcendentals $6^{th}$ Edition by James Stewart). For examples:
The function $y = f(x) = x^3$ is a one-to-one function and its inverse function is $y = f^{-1}(x) = x^{1/3}$. The function $y = f(x) = x^2 -2$ is not a one-to-one function b/c if we solve for $x$: $x = ±\sqrt{y+2}$ or $y =  ±\sqrt{x+2}$ which are two different functions of $x$. The function, $y = f(x) = x^2 - 2$, itself, does not pass the Horizontal line test to be one-to-one function.(see Ibid)
