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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?

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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.

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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 $C=0$).

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