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Can anyone help me to solve this differential equation. Obviously this is not a linear differential equation. So the only way that I know to solve non linear differential equations is by "manipulating" the $y$ variable in order to become a linear differential equation. But in this case I cant find a way. My main problem is that there is a $dy/dx$ inside and an $\ln$ outside .

\begin{align} \ y - \ln(\dot y) = x\dot y \ \end{align}

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    $\begingroup$ Use normal text for text which is not contained in a mathematical expression i.e. $\mbox{use this rarely!}$ , I'll change it this time, and sort out some English grammar. In any case, your attempt was well-directed (and wouldn't work here as I can vouch for as well) but you'll do well to remember what's written here. $\endgroup$ Commented May 9, 2022 at 6:50

2 Answers 2

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A differential equation of the form $y(x)=xy'+f(y')$ is called a Clairaut equation. To solve Clairaut's equation, one first differentiates with respect to $x$. In this case,

$$y'=xy''+y'+\frac{y''}{y'}=y'+y''\left(x+\frac{1}{y'}\right).$$

Which implies

$$y''\left(x+\frac{1}{y'}\right)=0.$$

In Clairaut's equation, either $y''=0$ or $x+f'(y')=0$.

For $y''=0$,

$$y'=\int0dx=C\implies y=\ln(C)+Cx.$$

For $x+\frac{1}{y'}=0$,

$$y'=-\frac{1}{x}\implies y=\ln\left(-\frac{1}{x}\right)-1.$$

That is,

$$y=\ln\left(-\frac{1}{x}\right)-1\wedge y=\ln(C)+Cx.$$

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This can be solved by maxima using the contrib_ode package. The equation eq is defined at line (%i2) and solved at line (%i3). Two solutions are returned as a list at line (%o3).

The solution method is given in the variable method as clairault, indicating it was solved using the Clairaut method.

(%i1) load('contrib_ode)$
(%i2) eq:y-log('diff(y,x))=x*'diff(y,x);
                                      dy      dy
(%o2)                         y - log(--) = x --
                                      dx      dx
(%i3) contrib_ode(eq,y,x);
                                      dy      dy
(%t3)                         y - log(--) = x --
                                      dx      dx

                     first order equation not linear in y'

                                                        %t x + 1
(%o3)   [y = %c x + log(%c), [y - %t x - log(%t) = 0, - -------- = 0]]
                                                           %t
(%i4) method;
(%o4)                              clairault

The two solutions are returned as a list.

The first solution y = %c x + log(%c) has a integration constant %c

The second solution

                             %t x + 1
  [y - %t x - log(%t) = 0, - -------- = 0]]
                                %t

is a parametric solution with parameter %t. It is a singular solution as it doesn't have an integration constant %c. The parameter %t can be eliminated with a little algebra: first solve for %t in terms of x (%o7), then substitute this result to obtain (%o8).

(%i6) soln[2][2];
                                  %t x + 1
(%o6)                           - -------- = 0
                                     %t
(%i7) solve(soln[2][2],%t);
                                          1
(%o7)                             [%t = - -]
                                          x
(%i8) subst(%,soln[2][1]);
                                       1
(%o8)                        y - log(- -) + 1 = 0
                                       x
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  • $\begingroup$ As it’s currently written, your answer is unclear. Please edit to add additional details that will help others understand how this addresses the question asked. You can find more information on how to write good answers in the help center. $\endgroup$
    – Community Bot
    Commented May 9, 2022 at 6:35
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    $\begingroup$ Your solution does not help readers understand how such problems are solved. In particular, it is not clear how Maxima solves them. Take a look at the other answers: you will see that they offer first some general knowledge about this type of equations, and then proceed to solving the given one. This is what one expects from a good answer, not solutions spit by a computer algebra system. $\endgroup$
    – Alex M.
    Commented May 9, 2022 at 11:25
  • $\begingroup$ Thanks for the feedback. I can't improve on the other answers, so I will leave it as is. $\endgroup$ Commented May 10, 2022 at 0:13

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