non linear differential equation that i cant solve

Question

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}

Answer 1

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,

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

Answer 2

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