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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?

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  • $\begingroup$ fixed missing dy $\endgroup$
    – Celeritas
    Feb 20, 2015 at 8:10
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    $\begingroup$ Solve for $x$ when the ODE looks (is) simpler than when solving for $y$. $\endgroup$ Feb 20, 2015 at 8:14
  • $\begingroup$ @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 $\endgroup$
    – Celeritas
    Feb 20, 2015 at 8:21
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    $\begingroup$ 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. $\endgroup$
    – abel
    Feb 20, 2015 at 20:37

3 Answers 3

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

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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.

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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)

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