The general form of the exact nonlinear ODE is given by the following:

exact_nonlinear_ode := 'diff(F(x,y(x),seq(diff(y(x),x$i),i=1..n)),x)' = 0;

$\mathrm{exact\_nonlinear\_ode}≔\frac{\partial }{\partial x}F\left(x\,y\left(x\right)\,\mathrm{seq}\left(\frac{{\ⅆ}^i}{\ⅆx^i}y\left(x\right)\,i=1..n\right)\right)=0$

See Murphy, "Ordinary Differential Equations and their Solutions", p. 221.

The order of this ODE can be reduced since it is the total derivative of an ODE of one order lower. If the given ODE is G(x,y,y1,y2,...,yn)=0, the test for exactness is the following:

where

Note: The derivatives with respect to y, dy/dx and d^2y/dx^2 are taken in the obvious manner but the derivatives with regard to x are taken considering y, and its derivatives as functions of x.

The reduced ODE is:

reduced_ode := 'F(x,y(x),seq(diff(y(x),x$i),i=1..n))' = _C1;

$\mathrm{reduced\_ode}≔F\left(x\,y\left(x\right)\,\mathrm{seq}\left(\frac{{\ⅆ}^i}{\ⅆx^i}y\left(x\right)\,i=1..n\right)\right)=\mathrm{\_C1}$

$\mathrm{with}\left(\mathrm{DEtools}\,\mathrm{odeadvisor}\right)$

$\left[\mathrm{odeadvisor}\right]$

$\mathrm{ode}≔\mathrm{diff}\left(y\left(x\right)\,x\,x\right)=\frac{1}{y\left(x\right)}-\frac{x}{{y\left(x\right)}^2}\mathrm{diff}\left(y\left(x\right)\,x\right)$

$\mathrm{ode}≔\frac{{\ⅆ}^2}{\ⅆx^2}y\left(x\right)=\frac{1}{y\left(x\right)}-\frac{x\left(\frac{\ⅆ}{\ⅆx}y\left(x\right)\right)}{{y\left(x\right)}^2}$

$\mathrm{odeadvisor}\left(\mathrm{ode}\right)$

$\left[\left[\mathrm{\_2nd\_order}\,\mathrm{\_exact}\,\mathrm{\_nonlinear}\right]\,\left[\mathrm{\_2nd\_order}\,\mathrm{\_with\_linear\_symmetries}\right]\,\left[\mathrm{\_2nd\_order}\,\mathrm{\_reducible}\,\mathrm{\_mu\_x\_y1}\right]\,\left[\mathrm{\_2nd\_order}\,\mathrm{\_reducible}\,\mathrm{\_mu\_y\_y1}\right]\,\left[\mathrm{\_2nd\_order}\,\mathrm{\_reducible}\,\mathrm{\_mu\_xy}\right]\right]$

$\mathrm{ans}≔\mathrm{dsolve}\left(\mathrm{ode}\,\mathrm{implicit}\right)$

$\mathrm{ans}≔-\frac{\ln \left(\frac{\mathrm{c\_\_1}xy\left(x\right)-x^2+{y\left(x\right)}^2}{x^2}\right)}{2}-\frac{\mathrm{c\_\_1}\mathrm{arctanh}\left(\frac{\mathrm{c\_\_1}x+2y\left(x\right)}{x\sqrt{{\mathrm{c\_\_1}}^2+4}}\right)}{\sqrt{{\mathrm{c\_\_1}}^2+4}}-\ln \left(x\right)-\mathrm{c\_\_2}=0$

$\mathrm{odetest}\left(\mathrm{ans}\,\mathrm{ode}\right)$

$0$