ODE Problem - how to proceed I am trying to solve this ODE:
$\dot{x}(t+x^2) = x$
...using this hint:
$\frac{dx}{dt} = \frac{1}{\frac{dt}{dx}}$.
I'm not entirely sure what this hint even means but I am guessing it has something to do with the inverse function. I would appreciate some guidance on the matter or at least an explanation of what this hint is supposed to mean.
 A: Suppose that you look for $t$ as a function of $x$m then $$
\frac{dt}{dx} = \frac{t+x^2}x
= \frac tx + x
$$
whose general solution is $$
t(x) = Cx + x^2
$$
[via variation a the constant: solution of $\frac{dt}{dx} = \frac tx$ is
$t(x) = Cx$, and variation a the constant gives $xC'(x) = x$ giving $
C(x) = x $ as a particular solution].
Now solve the equation in $x(t)$.
Suppose for example an initial condition $x(0) = x_0$.
$$
0 = t(x_0) = Cx_0 + x_0^2
$$
has two solutions: $x_0 = 0$ leads to the solution, otherwise $C = -x_0$ and
$$
t = -x_0x(t) + x(t)^2
$$
that is, $$
x(t) = \frac 12 (x_0 +\sqrt{x_0^2 + 4t})
$$ (the choice is done thanks to the initial condition).
A: $\dot{x}(t+x^2) = x$
$\frac{dx}{dt}(t+x^2) = x$
$tdx+x^2{dx}= xdt$
$tdx-xdt=-x^2{dt}$
$\frac{xdt-tdx}{x^2}=dx$
$d(\frac{t}{x})=dx$
$\int d(\frac{t}{x})=\int dx$
$\frac{t}{x}=x+C$
$t=x^2+Cx$ we can write this in $x^2+Cx-t=0$ form.Then we can solve this as quadratic equation.So
$D=C^2+4t$  then  $x=\frac{-C\pm \sqrt{C^2+4t}}{2}=-\frac{C}{2} \pm$ $\sqrt{(\frac{C}{2})^2+t^2}$
Finally solution of the equation is $x=-\frac{C}{2} \pm$ $\sqrt{(\frac{C}{2})^2+t^2}$
Note. $C$ is invariable
