Question: "How to solvey″+(1−y)(1+y)y=0"

Jackson Floyd

Answered question

2022-03-30

How to solve

y^{″} + (1 - y) (1 + y) y = 0

Cody Hart

Beginner2022-03-31Added 11 answers

Step 1

y^{″} + (1 - y) (1 + y) y = 0

Swithch variables to get

- \frac{x^{″}}{{[x^{'}]}^{3}} + (1 - y) (1 + y) y = 0

Reduction of order

p = x^{'}

gives

p = x^{'} = \pm \frac{1}{\sqrt{\frac{y^{4}}{2} - y^{2} + c_{1}}}

and now, you will face an ugly ellptic integral.

x + c_{2} = \pm \int \frac{d y}{\sqrt{\frac{y^{4}}{2} - y^{2} + c_{1}}}

Rewrite

\sqrt{\frac{y^{4}}{2} - y^{2} + c_{1}}

= \frac{1}{\sqrt{2}} \sqrt{(y^{2} - a) (y^{2} - b)}

where a and b are the roots of the quadratic in

y^{2}

and, hoping that I am not mistaken,

x + c_{2} = \pm \sqrt{\frac{2}{b}} F (\sin^{- 1} (\frac{y}{\sqrt{a}}) ∣ \frac{a}{b})

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