Is there a proper way of solving this differential equation of the second order? $$ \frac{d^2y}{dx^2}=ay^2 $$ Is it even possible?

  • $\begingroup$ Mathematica gives :$$y(x) = \left(\dfrac{6}{a}\right)^{3/2}\wp\left[ \left(\dfrac{a}{6}\right)^{3/2}\ (x+c_1); 0, c_2\right]$$ $\endgroup$ – Joelafrite Jun 13 '15 at 22:54
  • $\begingroup$ You shouldn't post the same question twice. Instead, just edit the original. $\endgroup$ – GPerez Jun 13 '15 at 22:57

(I'm going to set $y'=dy/dx$ and so on for brevity.) Multiply both sides by $2y'$: $$ 2y'y'' = 2ay'y^2. $$ Because $$ (y'^2)' = 2y'y'', $$ we can integrate the equation once to get $$ y'^2 = \frac{2a}{3}(y^3+A), $$ $A$ an arbitrary constant. Rearranging this and taking the square root, $$ \sqrt{2a/3}(x+B) = \pm \int \frac{dy}{\sqrt{ y^3+A }} $$ Except in special cases ($A=0$ being the only one, I suspect), the integral on the right cannot be solved in terms of elementary functions: a solution to this sort of integral equation (or the previous differential equation) is called an elliptic function (note the equations at the bottom of the page), because such things originate in finding the inverse function of the arc length of an ellipse.

  • $\begingroup$ according to the result of mathematica, it should be possible to say something for the integral you found. $\endgroup$ – Seyhmus Güngören Jun 13 '15 at 22:58

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