Are there any 'basic' solutions to this differential equation (ie using polynomials, exponetials, trigonometric functions and logarithms)? I cannot figure it out at all using the techniques I know for solving differential equations.
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Try $y=\frac {c}{x^2}$. – lulu Aug 29 '15 at 14:14
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Ah yes, I just realised that $y=6x^{-2}$ works but is there a more general solution? – bnosnehpets Aug 29 '15 at 14:16
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3To do it in general, let $z=y'$. Then $\frac {dy}{dz}=\frac {z}{y^2}$ which you can separate. – lulu Aug 29 '15 at 14:17
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Multiplying this equation by $2y'$, we get $$\left(y'^2\right)'=\frac23\left(y^3\right)',$$ which implies that $$y'^2=\frac23y^3+C_1.$$ Although the general solution of this can be written in terms of elliptic integrals (the equation is separable), there is a particularly simple solution corresponding to $C_1=0$: then $$y'=\pm \sqrt{\frac23}y^{\frac32}\qquad \Longrightarrow \qquad -2y^{-\frac12}=\pm \sqrt{\frac23} x+C_2.$$
Start wearing purple
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$$2y''y'=2y^2y'$$ $$(y')^2=\frac{2}{3}y^3+c_1$$ $$1=\frac{y'}{\sqrt{\frac{2}{3}y^3+c_1}}$$ $$x=\int\frac{dy}{\sqrt{\frac{2}{3}y^3+c_1}}+c_2$$ This elliptic integral cannot be expressed with a finite number of elementary functions. It can be reduced to an elliptic integral of the first kind, on the form of a complicated function $x(y)$. Furthermore, the inverse function $y(x)$ is not easy to derive.
It is better to refer to a particular case of Weierstrass's ODE : $$(y')^2=\frac{2}{3}y^3+c_1$$ http://mathworld.wolfram.com/WeierstrassEllipticFunction.html Eq.(34) where $g_2=0$
So, the simplest closed form of the general solution of $y''=y^2$ can be expressed on the form : $$y(x)=6^{1/3}\wp \left(\frac{x+C_2}{6^{1/3} };0\: ,\: C_1\right)$$ $\wp(\:)$ is the Weierstrass-p function.
JJacquelin
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A physicist's substitution always helps when the independent variable is missing and you have second derivatives. Let $t$ be the independent variable ($y'=dy/dt$). Then use a new variable $$y'=v$$ and substitute $$y''=v\,dv/dy$$ and integrate with respect to $y$.
In this case: $$v\,dv/dy=y^2$$ $$\int v\, dv=\int y^2 \, dy$$ $$v^2/2=y^3/3+C$$ Now put $v=dy/dt$ back in and integrate again to get $y(t)$.
In physics, you often have second derivatives over time (2nd Newton's law), so this is a routine bread-and-butter method for converting equations to first order and solving them routinely.
orion
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Alternative hint...note that $$y''=y'\frac{dy'}{dy},$$ which will make a separable variable DE
David Quinn
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