Is there a solution for this non-linear ODE involving exponentials?

Question

There is an non-linear ODE that pops out of the equations a lot when trying to solve the linear case of some second order ODE's. The equation is this: $$\ddot{y}+\dot{y}^2=y^2$$ It's easy to see that, for any function $y=f(x)$, we can multiply the equation by the integrating factor $e^{y}$ to get: $$e^{y}(\ddot{y}+\dot{y}^2)=y^2e^{y}$$ $$\frac{d^2}{dx^2}\left(e^{y}\right)=y^2e^{y}$$ Or $$e^{y}=\int\int y^2e^{y}dx^2$$ But, this seems like a dead end to me... Any help would be very much appreciated. Thanks in advance.

Answer 1

The ODE can be reduced to a quadrature. Notice that $\ddot{y}=\dot{y}\frac{d\dot{y}}{dy}=\frac{1}{2}\frac{d}{dy}\dot{y}^2$, so we can rewrite $\ddot{y}+\dot{y}^2=y^2$ as the first order linear ODE $$ \frac{1}{2}\frac{du}{dy}+u=y^2, \tag{1} $$ where $u=\dot{y}^2$. To solve it, multiply both sides of $(1)$ by $2e^{2y}$ and integrate: $$ e^{2y}\left(\frac{du}{dy}+2u\right)=\frac{d}{dy}\left(e^{2y}u\right)=2y^2e^{2y} $$ $$ \implies u=2e^{-2y}\int y^2e^{2y}dy=Ce^{-2y}+y^2-y+\frac{1}{2}. \tag{2} $$ From $(2)$ and the definition of $u$ it follows that $$ x-x_0=\pm\int_{y_0}^y\frac{d\xi}{\sqrt{Ce^{-2\xi}+{\xi}^2-\xi+\frac{1}{2}}}. \tag{3} $$

Answer 2

If you switch variables, you have $$-\frac{x''(y)}{[x'(y)]^3}+\frac{1}{[x'(y)]^2}=y^2$$ Reduction of order $p(y)=x'(y)$ gives $$-\frac{p'(y)}{[p(y)]^3}+\frac{1}{[p(y)]^2}=y^2$$that is to say $$p(y)=\pm\frac{e^y}{\sqrt{e^{2 y} \left((y-1) y+\frac{1}{2}\right)+c_1}}$$

Answer 3

Actually, i found an even more compact way to represent the solution, and it only uses the fact given by @Gonçalo that for any doubly differntiable function $y$, $\ddot{y}=\frac{1}{2}\frac{d}{dy}\dot{y}^2$. The solution can be obtained by starting with one equation that I posted in the original question body. For simplicity of notation, we'll define the $k$-th derivative wrt $f$ to be $d^k_f=\frac{d^k}{df^k}$ $$d_x^2e^y=y^2e^y$$ Which we just saw that is $$\frac{d_{e^y}(d_xe^y)^2}{2}=y^2e^y$$ By integrating wrt $e^y$, we obtain a Riemann-Steltjess integral: \begin{equation}(d_xe^y)^2=\dot{y}^2e^{2y}=2\int y^2e^yde^y\quad\quad(1)\end{equation} By RS integral properties, we can obtain $$2\int y^2e^yde^y=2\int y^2e^y(y'e^y)dx=2\int y^2e^{2y}dy$$ Substituing back into (1): $$\dot{y}^2e^{2y}=2\int y^2e^{2y}dy$$ Proceeding by integrating by parts $$\dot{y}^2e^{2y}=e^{2y}(y^2-y+\frac{1}{2})$$ Wich simplifies to: $$\dot{y}^2=y^2-y+\frac{1}{2}$$ The equation has two solutions: $$y_1(x)=\frac{1}{2}(1-\sinh(\sqrt{2}c+x))$$ $$y_2(x)=\frac{1}{2}(1+\sinh(\sqrt{2}c+x))$$