How to solve $x=tx'^2+x'^3$ How about the differential equation $x=tx'^2+x'^3$. Can somebody solve it?
my solution:suppose $\dot x$=p,then $x=tp^2+p^3$. 
Then, p=$p^2$+2tp$\dot p$+3$p^2\dot p$, then we can use these function to solve p, but I just find the final result is so complex. I don't know how to continue? Or, does somebody have other kinds of methods? Thank you
 A: Hint: You have:
$$p=p^2+2tp' p+3p^2p'$$
Divide by $p$, yielding:
$$1 - p - p'(2t - 3p)=0$$
Try to solve this using an exact form. It is not exact, so find an integrating factor to make it exact.
You will make it exact and find a somewhat ugly solution for $p$ in terms of a constant and then substitute back in the original $p = x'$.
A: A parametric solution isn't that horrible. 
If $x(t)$ is not constant, then $p(t) = \dot{x}(t)$ isn't identically zero and one can rewrite your equation as
$$\frac{dp}{dt} = \frac{1-p}{2t + 3p}\tag{*1} = -\frac{\tilde{p}}{2\tilde{t}+3\tilde{p}}
\quad\text{ where }\quad
\begin{cases}\tilde{p} &= p - 1\\\tilde{t} &= t+\frac32\end{cases}
$$
Introduce an auxiliary parameter $s$ such that
$$\frac{dt}{ds} = \frac{d\tilde{t}}{ds} = 2\tilde{t}(s) + 3\tilde{p}(s)$$
Equation $(*1)$ can be recasted into a matrix form:
$$\frac{d}{ds} \begin{bmatrix}\tilde{p}(s)\\\tilde{t}(s)\end{bmatrix} = 
\begin{bmatrix}-1 & 0\\3 & 2\end{bmatrix} \begin{bmatrix}\tilde{p}(s)\\\tilde{t}(s)\end{bmatrix}
$$
The matrix 
$\begin{bmatrix}-1 & 0\\3 & 2\end{bmatrix}$ has eigenvalues $-1$ and $2$ with corresponding eigenvectors $\begin{bmatrix}1\\-1\end{bmatrix}$ and $\begin{bmatrix}0\\1\end{bmatrix}$. 
From this, we find
$$
\begin{bmatrix}p(s)\\t(s)\end{bmatrix} =
\begin{bmatrix}1\\{\small -\frac32}\end{bmatrix} 
+ Ae^{-s} \begin{bmatrix}1\\-1\end{bmatrix}
+ \frac{B}{2}e^{2s} \begin{bmatrix}0\\1\end{bmatrix}
\tag{*2}
$$
for suitable chosen constant $A$ and $B$. If $p(t)$ is not the constant function $1$, then $A \ne 0$ and we can absorb it into $e^{-s}$ and set it to $\text{sign}(A) = \pm 1$. After suitable rescaling of $B$ and let $u = \mp e^{-s}$, we can transform $(*2)$ as
$$
\begin{cases}
p(u) &= 1 - u\\
t(u) &= -\frac32+ u + \frac{B}{2u^2}
\end{cases}
$$
As a function of $u$, we now have
$$
\frac{d x(u)}{du} = \frac{d x(t)}{dt} \frac{dt(u)}{du} = (1 - u)\left(1 - \frac{B}{u^3}\right)
= 1 - u + \frac{B}{u^2} - \frac{B}{u^3}
$$
and hence
$$
\begin{cases}
x(u) &= u - \frac{u^2}{2} - \frac{B}{u} + \frac{B}{2u^2} + C\\
t(u) &= -\frac32 + u + \frac{B}{2u^2}
\end{cases}\tag{*3}
$$
where $C$ is yet another integration constant. 
Substitute $(*3)$ into the expression 
$x(t) - tx'(t)^2 - x'(t)^3$, we find it equal to the constant $C - \frac{B-1}{2}$. 
This fixes $C$ to $\frac{B-1}{2}$ and the original ODE has parametric solutions of the form:
$$
\begin{cases}
x(u) &= u - \frac{u^2}{2} - \frac{B}{u} + \frac{B}{2u^2} + \frac{B-1}{2}
= \frac{(B-u^2)(1-u)^2}{2u^2}\\
t(u) &= -\frac32 + u + \frac{B}{2u^2}
\end{cases}
$$
