Solve $\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$

Question

Solve the equation, $$\frac{d^2y}{dx^2}+y=\frac{1}{y^3}$$ We have $$y^3\frac{d^2y}{dx^2}+y^4=1$$ I tried using change of dependent variable

Let $z=y^3\frac{dy}{dx}$ Then we get $$y^3\frac{d^2y}{dx^2}+3y^2\left(\frac{dy}{dx}\right)^2=\frac{dz}{dx}$$ But i could not get an equation completely involving $z,x$

Answer 1

HINT

\begin{align*} y'' + y = \frac{1}{y^{3}} & \Longleftrightarrow y''y' + yy' = \frac{y'}{y^{3}}\\\\ & \Longleftrightarrow (y')^{2} + y^{2} = -\frac{1}{y^{2}} + c\\\\ & \Longleftrightarrow (y')^{2} = \frac{cy^{2} - y^{4} - 1}{y^{2}}\\\\ & \Longleftrightarrow y' = \pm\sqrt{\frac{cy^{2} - y^{4} - 1}{y^{2}}} \end{align*}

Answer 2

Another Way to look at it

$$\frac{d^2y}{d^2x} = \frac{1}{y^3} - y => \frac{d^2y}{d^2x} = \frac{1-y^4}{y^3} =>$$ $$\frac{y^3}{1-y^4} d^2y = d^2x $$

Now this can be double integrated using the integral of a division formula

Answer 3

Another possibility

Switch variables to make $$-\frac {x''}{[x']^3}+y=\frac 1 {y^3}$$ Reduction of order $p=x'$ gives a separable equation $$-\frac {p'}{p^3}=\frac 1 {y^3}-y\implies p=\pm\frac{y}{\sqrt{c_1 y^2-y^4-1}}$$ Now $$x+c_2=\pm \int \frac{y}{\sqrt{c_1 y^2-y^4-1}}\,dy=\pm \frac 12 \int \frac{dt}{\sqrt{c_1 t-t^2-1}}$$ and you will arrive to some arctangent.