Implicit solution to second-order non-linear differential equation I am trying to solve the following differential equation:
$$
\frac{8}{9}y(t) - 2y(t)^2 + y(t)^3 -y''(t) =0.
$$
I think it is impossible to solve it for $y(t)$, but apparently it is possible to solve it for $t(y)$. I really do not have a clue how to solve this one. I guess I need to do some kind of transformation, but I don't really know which one. Can anyone help me?
 A: Multiply the equation by $2y'$, this will transform it into
$$\frac{d}{dt}\left(\frac89y^2-\frac43y^3+\frac12y^4-y'^2\right)=0.$$
Using this first integral and separable form of the resulting 1st order ODE, we can write the implicit solution in quadratures:
$$\int\frac{dy}{\sqrt{C_1+\frac89y^2-\frac43y^3+\frac12y^4}}=\pm t+C_2.$$
The integral on the left is in principle expressible in terms of elliptic functions.
A: Assuming that $y^{\prime}(t)\neq0$ at some $t$, it follows by the continuity
of $y^{\prime}$ that this is valid for an entire interval, i.e., there exists
an interval $I$ such that $y^{\prime}(t)\neq0$ for all $t\in I$. This means
that $y^{\prime}$ has constant sign (being continuous), hence $y$ is strictly
monotone on $I$, hence $y:I\rightarrow J=y(I)$ is invertible and for
convenience denote the inverse of $y$ with $t$ (a slight abuse of notation).
In this way
$$
t\in I\overset{y}{\longmapsto}y=y(t)\in J
$$
and
$$
y\in J\overset{t}{\longmapsto}t=t(y)\in I.
$$
Now, since
$$
\left(  t\circ y\right)  (t)=t\text{  for all }t\in I
$$
it follows that
$$
\left(  t\circ y\right)  ^{\prime}(t)=1\text{  for all }t\in I
$$
meaning that
$$
y^{\prime}(t)\cdot t^{\prime}(y(t))=1
$$
or, for short,
$$
y^{\prime}(t)=\frac{1}{t^{\prime}(y)} \text{for all }t\in I\text{.}
$$
Next,
$$
\left(  y^{\prime}\cdot(t^{\prime}\circ y)\right)  ^{\prime}(t)=0
\text{for all }t\in I
$$
which leads to
$$
y^{\prime\prime}(t)\cdot t^{\prime}(y(t))+y^{\prime}(t)\cdot y^{\prime
}(t)\cdot t^{\prime\prime}(y(t))=0
$$
hence
$$
y^{\prime\prime}(t)=-\frac{\left(  y^{\prime}(t)\right)  ^{2}\cdot
t^{\prime\prime}(y)}{t^{\prime}(y)}=-\frac{t^{\prime\prime}(y)}{\left(
t^{\prime}(y)\right)  ^{3}}\text{.}
$$
Coming back to the equation
$$
\frac{8}{9}y(t)-2y(t)^{2}+y(t)^{3}-y^{\prime\prime}(t)=0
$$
and considering $y$ to be the variable and $t$ to be the function (i.e.,
$t=t(y)$), it follows that
$$
\frac{8}{9}y-2y^{2}+y^{3}=-\frac{t^{\prime\prime}(y)}{\left(  t^{\prime
}(y)\right)  ^{3}}
$$
or
$$
t^{\prime\prime}+\left(  y^{3}-2y^{2}+\frac{8}{9}y\right)  \left(  t^{\prime
}\right)  ^{3}=0.
$$
In order to solve this equation, let $u:=t^{\prime}$, hence
$$
u^{\prime}+\left(  y^{3}-2y^{2}+\frac{8}{9}y\right)  u^{3}=0
$$
which is a separable equation. Clearly $u\neq0$, since $t^{\prime}\neq0$, so
we can divide with $u^{3}$ and obtain
$$
-u^{\prime}u^{-3}=y^{3}-2y^{2}+\frac{8}{9}y
$$
which we integrate (w.r.t. $y$) and get
$$
\frac{1}{2}u^{-2}=\int\left(  y^{3}-2y^{2}+\frac{8}{9}y\right)  dy=\frac{1}
{4}y^{4}-\frac{2}{3}y^{3}+\frac{4}{9}y^{2}+C=\left(  \frac{y(3y-4)}{6}\right)
^{2}+C\text{, }C\in\mathbb{R}
$$
which leads to
$$
u=\frac{1}{\sqrt{2\left(  \dfrac{y(3y-4)}{6}\right)  ^{2}+C}}
$$
hence
$$
t(y)=\int\frac{dy}{\sqrt{2\left(  \dfrac{y(3y-4)}{6}\right)  ^{2}+C}}.
$$
In this way you can express $t$ as a function of $y$ and get an implicit form
of the solution $y$. Unfortunately, I don't know if there is a way of
computing the integral (except for $C=0$).
