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?

• fixed missing dy Feb 20, 2015 at 8:10
• Solve for $x$ when the ODE looks (is) simpler than when solving for $y$. Feb 20, 2015 at 8:14
• @ClaudeLeibovici so when deciding which method you are going to use to solve the ODE, you consider both solving for x and y? So what I'm saying is there's two things to consider 1) which method to use to solve ODE 2) which variable to solve for Feb 20, 2015 at 8:21
• at most points the relation between $x$ and $y$ is reversible. that is $\frac{dy}{dx}\, \frac{dx}{dy} = 1.$ in other words $y(x)$ and $x(y)$ inverses of each other.
– abel
Feb 20, 2015 at 20:37

$$y\frac{dx}{dy}-4(x+y^6)=0$$ 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 $$2t^3+ct^2-x=0$$

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.

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)