I'm trying to solve the first-order ordinary differential equation $x \cdot y'(x)-y(x)=log(y'(x))$.

I've noticed that the equation is separable, but I'm not able to solve the indefinite integral $\int \frac{log(y'(x)) + y(x)}{y'(x)} dx$. I've tried to find a substitution that would help me solve the equation (for example I've checked $log(y'(x)), log(y(x))$ ) but none has worked.

Also, by differentiating both sides I get $y'(x)=\frac{1}{x}$, and thus I've found a private solution $y(x)=log(x)+1$. However, I haven't managed finding a general solution using the private solution (writing $y(x)=log(x)+1+u(x)$ doesn't work).

Another approach that I've considered is approximating each solution (by considering a variable starting point) using Picard's or Euler's method up to some order, finding the general case and using the convergence theorem of the method in order to find the general solution. However, since I cannot write $y'(x)=g(x,y)$ for some $g$, it doesn't seem possible here.

What else may I do in order to solve this ODE?

Thanks in advance.


2 Answers 2


Actually, differentiating gives you $$ x y'' = \frac{y''}{y'}$$ If $y'' \ne 0$ this says $y' = 1/x$, but the other possibility is $y'' = 0$, which actually leads to the general solution: $$y = c x - \ln(c)$$

  • 1
    $\begingroup$ So $log(x)+1$ is a singular solution? $\endgroup$
    – RanSch
    Feb 4, 2019 at 13:25

Your equation has the form of a Clairaut DE, $$ y=xy'+f(y') $$ with here $f(v)=-\ln v$. This type of equation has a solution family of straight lines $$ y=cx+f(c)=cx-\ln c $$ and an envelope given by $0=x+f'(y')$, here leading to $y'=\frac1x$ so that $$ y = x\frac{1}{x}-\ln(\frac1x)=1+\ln(x)~~\text{ for }~~x>0. $$

Where the lines touch the envelope, further solutions can be constructed as a piecewise functions.


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