# How to solve this second order differential equation

• I
• Safinaz
This can be solved by radicals:\begin{align*}ae^y+2c&=k\sqrt{ae^y+be^{2y}+c} \\ae^y+2c&=k(\sqrt{ae^y}) \\&=k(\sqrt{1+e^x})\end{align*}This last equation gives the solution for y as:$$y=-\frac{\tanh(\sqrt{c}(x+c_2))}{\sqrt{c}}f #### Safinaz Any idea how to solve this equation: ## \ddot \sigma - p e^\sigma - q e^{2\sigma} =0 ## Or ## \frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0 ## Where p and q are constants. Thanks. Delta2 This is of the form $\ddot \sigma = f(\sigma)$, so you can reduce it to first order by multiplying both sides by $\dot \sigma$ and integrating: $$\frac12 \left(\dot\sigma^2(t) - \dot\sigma^2(0)\right) = \int_{\sigma(0)}^{\sigma(t)} f(s)\,ds.$$ I haven't checked whether the resulting ODE for $\dot \sigma$ can be solved analytically. Delta2 The equation is second order, all right, but it is not a partial differential equation, as you initially wrote in the thread title. It's a second order ordinary differential equation (ODE). Last edited: Any idea how to solve this equation: ## \ddot \sigma - p e^\sigma - q e^{2\sigma} =0 ## Or ## \frac{d^2 \sigma}{dt^2} - p e^\sigma - q e^{2\sigma} =0 ## Where p and q are constants. Thanks. Here's how things go in the era of AI of the 21st century: You "feed" this ODE to Mathematica on the Wolframalpha website. In a matter of seconds, you will see if it has a solution in terms of known elementary or special functions. If such a solution is shown by the software, then you can ask people how this solution is computed by the program. If the AI can't break it, neither can people, then a series-expansion is needed (i.e. an approximation for t very close to 0). Here's how things go in the era of AI of the 21st century: You "feed" this ODE to Mathematica on the Wolframalpha website. In a matter of seconds, you will see if it has a solution in terms of known elementary or special functions. If such a solution is shown by the software, then you can ask people how this solution is computed by the program. If the AI can't break it, neither can people, then a series-expansion is needed (i.e. an approximation for t very close to 0). No. No. I am sorry, but if the OP has no "feeling" that his highly nonlinear ODE may be solved by special functions, he can't just pretend that AI did not exist, like in the time of Gradsheyn and Rytzhik doing complicated integrals by hand. You may think it is cheating, but I say this is only being being realistic. You can start with separation of variables and integrating one time ... It seems that the first integration will work, for the second I don't know, you must to try ... Ssnow We can interpret it as equation of motion $$\ddot{x}=-\frac{\partial U}{\partial x}$$ where $$U=-pe^x-\frac{q}{2}e^{2x}$$ We can investigate shape of the potential energy including $$U(-\infty)=0, |U(+\infty)|=\infty$$ Say p,q>0 total energy E<0 would tell the region where the particle cannot go. For E>0 the particle would go infinite. Last edited: (* edited: noticed DE in p easily solved by separation of variables *) I suggest start by writing it as$$
y''-a e^y-b e^{2y}=0
$$Then let ##y'=p## as the standard notation. Then ##y''=p\frac{dp}{dy}## leaving$$
p\frac{dp}{dy}-a e^y-b e^{2 y}=0
$$or$$
pdp=(ae^y+be^{2y})dy
$$It's at least reduced to first-order and can be solved by integrating:$$
p_1(y)=-\sqrt{2 a e^y+b e^{2 y}+2 c_1}

p_2(y)=\sqrt{2 a e^y+b e^{2 y}+2 c_1}
$$Then need to solve:$$
y'=-\sqrt{2 a e^y+b e^{2 y}+2 c_1}

y'=\sqrt{2 a e^y+b e^{2 y}+2 c_1}
$$which Mathematical gives solutions in terms of radicals, tanh, and log expressions: Last edited: anuttarasammyak would be interesting to figure out how these are the solutions. As said in post #8 these solutions come from energy conservation law. c_1 is total energy for m=1. Last edited: I'd like to continue a bit to solve this somewhat manually. We have the general expression:$$
\frac{dy}{dx}=\sqrt{a e^y+b e^{2y}+c}
$$Separating variables again and integrating (I did the integration with Mathematica):$$
\begin{align*}
\int\frac{dy}{\sqrt{a e^y+b e^{2y}+c}}&=x+c_2 \\
-\frac{\tanh ^{-1}\left(\frac{a e^y+2 c}{2 \sqrt{c} \sqrt{a e^y+b e^{2 y}+c}}\right)}{\sqrt{c}}&=x+c_2 \\
\frac{a e^y+2 c}{2 \sqrt{c} \sqrt{a e^y+b e^{2 y}+c}}&=\tanh(\sqrt{c}(x+c_2))
\end{align*}
$$So basically now reduced to an algebraic problem solving for y in the expression:$$
ae^y+2c=k\sqrt{ae^y+be^{2y}+c}


Last edited:
Ssnow