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How to solve the following differential equation:

$$ y' = \frac{\sqrt[3]{x-2y}}{\sin(x)} $$

We can rewrite it as

$$ y= \frac{1}{2} (x- y'^3 \sin^3(x))$$

I tried to substitute $y'=p$ and then proceed, but I did not conclude anything.

$\textbf{Edit:}$

I tried to rewrite it as:

$$ dy - \frac{\sqrt[3]{x-2y}}{\sin x} dx=0 $$ And then find the integrating factor:

$$ \mu = \mu (x) = e^{- \int \frac{\sqrt[3]{x-2y}}{\sin x}dx} $$

However there is no elementary way to solve the integral on the exponent.

$\textbf{Edit 2}$ I saw this Calculate Laurent series for $1/ \sin(z)$ So one way is perhaps to decompose the function:

$$ f(x) = \frac{1}{\sin(x)} $$

However I'm not familiar with solutions of DE with series and I don't know under what conditions we can do that.

Grant
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  • @Moo i tried to add my attempt, thank you for your suggestion. – Grant Nov 24 '23 at 22:38
  • Let $x-2y=u$... – Bob Dobbs Nov 25 '23 at 14:32
  • @BobDobbs Hmm, I don't see how we can recover it even if we make that substitution – Grant Nov 25 '23 at 14:35
  • Yes. I'm excited. Wolfram, I connect, couldn't solve. Maybe other applications can solve. – Bob Dobbs Nov 25 '23 at 14:45
  • @BobDobbs Maple couldn't either – Grant Nov 25 '23 at 15:13
  • Series method? Can you solve the DE if you replace $\sin x$ with x? – Bob Dobbs Nov 25 '23 at 15:46
  • Solving non-linear differential equations is considerably tougher than solving linear differential equations. Unless the differential equation itself was developed from a particular function, it is highly unlikely that it has a solution expressible in terms of elementary functions. It is curious that someone not even familiar with solving differential equations by power series - a basic technique for linear differential equations - would have cause to solve a non-linear equation. Where did this come from? – Paul Sinclair Nov 26 '23 at 03:29
  • @PaulSinclair it was on my homework. I was looking more for a confirmation that it cannot be solved with elementary methods. It's interesting that you're questioning one's curiosity! – Grant Nov 26 '23 at 15:04
  • I didn't question curiousity. Please do not exagerate what I said. I pointed out that this sort of problem is generally far beyond the level of the techniques you are employing or have expressed familiarity with. Thus it is surprising that you would have need to solve it. Integrating factors are a solution method for linear differential equations. Series expansions can be used for non-linear DEs, but are considerably harder to generate than they are for linear DEs. I could understand if it came from some practical application, but not as a homework assignment. – Paul Sinclair Nov 26 '23 at 16:02
  • The general focus for non-linear equations is to show under what conditions solutions exist, and are unique. Actually generating those solutions is not a significant concern, as there are various numerical techniques for calculating values of the function to some desired accuracy. I can only presume that either you are supposed to solve a linear approximation to this equation, (whose solution approximates the actual solution near some point), or else there is some specialized information in the chapter that is applicable to this case. – Paul Sinclair Nov 26 '23 at 16:09

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