How are second order nonlinear ordinary differential equations solved?

I conceived the following second order nonlinear ordinary differential equation:

$$\frac{d^2y(x)}{dx^2}=\frac{k}{(y(x))^2}$$

I can tell it's nonlinear because of the $\frac{k}{(y(x))^2}$ term and second order because of the second order derivative.

Also, I did some research and concluded that it is of the type "missing x". In this category we use the relation, according to SOS math : $$\frac{d^2y(x)}{dx^2}=v\frac{dv}{dy} (1)$$ by making v equal to $$\frac{dy(x)}{dx} (2)$$ and then, using the chain rule, simplify to obtain equation (1). Despite all that, I can't seem to find a logical solution to my ODE. I would appreciate any help or clue! Thanks!

• "Missing $x$" is called: autonomous differential equation.
– Did
Apr 4, 2016 at 13:59
• Are you sure sure the ODE in question is of that type? Apr 7, 2016 at 23:29
• ?? Trivially so, yes.
– Did
Apr 8, 2016 at 5:30

Typically, if your equation has a second derivative and a zeroth derivative but no first derivative, you can reduce the order by multiplying both sides by the first derivative and integrating. This works mathematically because the $y''$ gives you the "du" you need to integrate the left side, while the $y'$ gives you the "du" you need to integrate the right side. In many physical situations, this amounts to identifying a conservation law.

In this case you wind up with:

$$\frac{1}{2} y'^2=-\frac{k}{y}+C.$$

This equation is separable, provided you can unambiguously choose a sign for the square root.

• What is your source? And where can I find it (if there is one)? Apr 3, 2016 at 17:08
• @Investor Sorry to say that this wasn't taught in my elementary ODEs class, and I didn't have a book in the class where I did learn it. However, this is a standard technique in physics: $my''=f(y)$ so $my''y'=f(y)y'$ so $\frac{1}{2}y'^2-F(y)$ is constant. Thus you have identified the kinetic energy $\frac{1}{2} y'^2$ and the potential energy $-F(y)$ and demonstrated that the total energy is conserved. This works more generally in higher dimensions if you replace $f(y)$ with $\nabla F(y)$.
– Ian
Apr 3, 2016 at 17:37

You wrote that $\frac{d^2y}{dx^2}=v\frac{dv}{dy}$, so substitute that into your equation and you get $$v\frac{dv}{dy}=\frac{k}{y^2}.$$Now, $$v\ dv=\frac{k}{y^2}dy$$ $$-\frac{k}{y}+C_1=\frac{v^2}{2}$$ $$-\frac{2k}{y}+C_2=v^2$$ $$v=\pm\sqrt{-\frac{2k}{y}+C_2}$$Then, integrate and you get$$y = \pm\left(\sqrt{C_2-\dfrac{2k}{y}}y+\dfrac{k\ln\left(\frac{\left|\sqrt{C_2-\frac{2k}{y}}-\sqrt{C_2}\right|}{\sqrt{C_2-\frac{2k}{y}}+\sqrt{C_2}}\right)}{\sqrt{C_2}}\right)+C_3.$$Haven't checked yet, though... (Link here to the complicated integral, and click the "Go!" button.)