An Interesting Homemade Differential Equation

[@scottleung9587]

Nice!

[@JohnsonIdris112]

And you did forget to show us how you came up with the problem 😅

[@Hemicubesquare]

How about this?
actually
y'=+-(y^4-y^2+c)^1/2
if c=1/4
y'=+-(y^4-y^2+1/4)^1/2
=+-(y^2-1/2)^(2×1/2)
=+-(y^2-1/2)
dy/dx=+-(y^2-1/2)
if y=+-(1/2)^1/2
dy/dx=0
dy/+-(y^2-1/2)=dx
but
1/(y^2-1/2)+
=[1/{y-(1/2)^(1/2)}-1/{y+(1/2)^(1/2)}]2^(-1/2)
let's do integral about dy/dx
+-{ln|y-(1/2)^(1/2)|-ln|y+(1/2)^(1/2)|}2^(-1/2)
=+-ln|{y-(1/2)^(1/2)}/{y+(1/2)^(1/2)}|2^(-1/2)
=x+c
so
e^{+-(x+c)2^(1/2)}=|{y-(1/2)^(1/2)}/{y+(1/2)^(1/2)}|
=|{1-2^(1/2)/{y+(1/2)^(1/2)}|
if {1-2^(1/2)/{y+(1/2)^(1/2)}>0
y>(1/2)^(1/2) or y

[@davidblauyoutube]

At 7:15, substitute u = y^2+1/3 and simplify to obtain (u')^2 = 4u^3 + 4(2c1 -1/3)u + 4(2c1 /3 - 2/27) = 4 u^3 - g2 u - g3. Now if c1 = 0 or 1/8 then the original equation can be factored and a simpler solution can be found (as in the video and the other comments). These solutions correspond to g2^3 - 27 g3^2 = 0 (see the link below). Otherwise, the general solution to this differential equation is the Weierstrass elliptic function u = ℘(x), so y = ±√(℘(x)-1/3).
en.wikipedia.org/wiki/Weierstrass_elliptic_function

[@SyberMath]

Nice!

[@giuseppemalaguti435]

Posto dy/dx=u... l'equazione risulta (du/dy)u=2y^3-y...integro...u^2/2=2y^4/4-y^2/2+c..u=√(y^4-y^2+2c)...quindi X+C2=INT(dy/√(y^4-y^2+2c)...poi bisogna un po' lavorare..

[@Christopher-e7o]

X,2×+5=8

[@holyshit922]

F(y'',y'y) = 0
so it is not so difficult to solve
y' = u(y) is the standard substitution

[@Roq-stone]

Please don’t say, “multiply both sides by ‘dx’”,
It is integrate both sides with respect to x.

[@mcwulf25]

You could have put y=0 and made it simpler still 😂😂😂.
Thanks, nice trig solution.

[@SyberMath]

You're welcome!

[@Don-Ensley]

problem
d² y/dx² = 2 y³-y
Let
p = dy/dx
d² y/dx² = dp/dx
Use the chain rule.
d² y/dx² = (dp/dy)(dy/dx)
Replace dy/dx with p.
d² y/dx² = p (dp/dy)
Replace d² y/dx² in equation.
p (dp/dy) = 2 y³-y
Integrate.
∫ p dp = ∫ (2 y³-y) dy
p²/2 = 2 y⁴/ 4 - y²/ 2 + m
. Where m is a constant of integration.
p²/2 = y⁴/ 2 - y²/ 2 + m
p² = y⁴ - y² + 2m
p = ± √(y⁴ - y² + 2m)
= dy/dx
∫ dy / √(y⁴ - y² + 2m)
= ± ∫ dx
= c ± x
, where c is a constant of integration.
Set m = 0. The non-0 m version is too complicated for this channel.
∫ dy / √(y⁴ - y²) = c ± x
∫ dy / [ y √(y² - 1) ] = c ± x
Substitute
t = √(y² - 1)
dt = y dy / √(y² - 1)
dt/y = dy / √(y² - 1)
t² + 1 = y²
dy / [ y √(y² - 1) ]
= dt / y²
= dt / (1 + t²)
∫ dt / (1 + t²)
= c ± x
tan⁻¹ t = c ± x
t = tan (c ± x)
√(y² - 1) = tan (c ± x)
y² - 1 = tan² (c ± x)
y² = 1 + tan² (c ± x)
Use a trigonometric identity.
y² = sec² (c ± x)
y = ± sec (c ± x)
secθ = 1 / cosθ
It is even parity so sign does not commute to the argument. -cos(θ) cos(-θ)
answer
y ∈ { - sec (c - x),
- sec (c + x),
sec (c - x),
sec (c + x),
c ∈ ℝ }