# non linear differential equation that i cant solve

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}

• 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. Commented May 9, 2022 at 6:50

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,

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

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