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$$y-xy'=\exp(y')$$

I want to solve this differential equation, which looks simple but hard to solve. Any method?

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    $\begingroup$ Please use MathJax to typeset mathematics. $\endgroup$ Commented Mar 17, 2021 at 17:23
  • $\begingroup$ What makes it look easy ? I find it terrible ! $\endgroup$
    – user65203
    Commented Mar 17, 2021 at 17:25

1 Answer 1

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Use Clairaut equation to find the solution.

The general solution is $$y(x)=Cx+e^{C}.$$

Proof from Wikipedia article.

Differentiate both sides with respect to $x$ $$y'=y'+xy''+\exp\left(y'\right)y'',$$ so $$\left[x+\exp\left(y'\right)\right]y'' = 0.$$

General solution: $$y'' = 0 \Longrightarrow y(x)=Cx+e^C.$$

Singular solution: $$x+\exp\left(y'\right) = 0.$$


Thank you @zwim: Desmos calculator.

Clairaut equation

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    $\begingroup$ Is this solution unique? What about the singular solution? $\endgroup$
    – Nacib Neme
    Commented Mar 17, 2021 at 17:30
  • $\begingroup$ Can we solve a new differential equation, that is, y-x^2=exp(y'), using the same method? $\endgroup$ Commented Mar 17, 2021 at 17:46
  • $\begingroup$ Note that $y-x^2=\exp(y')$ is not a Clairaut equation. $\endgroup$
    – vitamin d
    Commented Mar 17, 2021 at 17:55
  • $\begingroup$ How should it be solved then? $\endgroup$ Commented Mar 17, 2021 at 17:56
  • $\begingroup$ which has a similar form $\endgroup$ Commented Mar 17, 2021 at 17:56

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