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/sci/ - Science & Math

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5773163 No.5773163 [Reply] [Original]

/sci/ I know you hate help threads but could someone help me with this?
My professor is being a bit of a dick and refusing to help anyone.

Its a nonlinear ODE:
<span class="math">\ddot y + 2y(\dot y)^3 = 0[/spoiler]
I made the substitution
<span class="math">\dot y = z[/spoiler]
so then
<span class="math">\ddot y = z\frac{dz}{dy} [/spoiler]
so the equation becomes sperable and linear
<span class="math">z\frac{dz}{dy} +2yz^3 = 0[/spoiler]

But I think I keep fucking up the integration, I end up with
<span class="math">x = \frac{1}{3}y^3 + cy + d[/spoiler]
Would someone mind checking my work to see where I messed up?

>> No.5773192

if you're attempting to troll, you really got me mad, 10/10.

>> No.5773202

Integrating factor method?

>> No.5773212

divide by <span class="math">(dy/dx)^2[/spoiler].

<span class="math">\frac{d^2y/dx^2}{(dy/dx)^2} +2 y dy/dx=\frac{d}{dx} (\frac{-1}{dy/dx} + y^2) = 0[/spoiler]

so <span class="math">y^2- 1/(dy/dx) [/spoiler] is a constant <span class="math">c[/spoiler](where the derivative isn't 0 obviously)

>> No.5773210

>>5773202
Oh shit neve mind ignore that

>> No.5773233

>>5773212
multiply then both sides by <span class="math">y'[/spoiler]:

<span class="math">y^2 y' - c y' = 1 [/spoiler]
so <span class="math"> (\frac{y^3}{3} - \frac{cy^2}{2})' = 1[/spoiler]
so there exists some <span class="math">d[/spoiler] such as
<span class="math"> \frac{1}{3} y^3 - \frac{1}{2} c y^2 = x + d [/spoiler]
cont

>> No.5773296

>>5773233
See this is whats tripping me up;
if you go from
<span class="math">\dot yy^2-\dot yc = 1[/spoiler]
and factor out the derivative you get
<span class="math">\dot y(y^2 - 1) = 1[/spoiler]
then doing the integral gives
<span class="math">\frac{1}{3}y^3+cy = x + d[/spoiler]
but if you keep doing it your way you end up with
<span class="math">\frac{1}{3} y^3 - \frac{1}{2} c y^2 = x + d
[/spoiler]

>> No.5773324 [DELETED] 

<span class="math">\begin{cases} \frac{du}{dx} =-2 y u^3 \\ \frac{dy}{dx} = u \end{cases}[/spoiler]
<span class="math">\implies \frac{du}{dy} = -2 y u^2[/spoiler]
<span class="math">\implies u = \frac{1}{y^2 + C_1}[/spoiler]
<span class="math">\implies x = \frac{1}{3} y^3 + C_1 y + C_2[/spoiler]