# How to solve the differential equation: $y-xy'=\exp(y')$ [closed]

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

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

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

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.

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