Equations with dependent variable missing

[Joseph1739]

Information from the book:
Use
1. v = y'
2. v' = y''
3. dy/dt = v
to solve the differential equation.

Question: 2t²y'' + (y')³ = 2ty'
I'm stuck at finding the integrating factor (which my book tells me is v^-3 in the solutions.)

Using the information above:
2t²v' + (v^3) - 2tv = 0
M(t,v) = v³ - 2tv
N(t,v) = 2t²
M_v = 3v² - 2t
N_t = 4t
Since these two are not equal, I have to find an integrating factor, but using both:
(1) dμ/dt = [(M_v-N_t)/(N)] * μ
(2) dμ/dv = [(N_t-M_v)/(M)] * μ
both result in a unsolvable equation because I end up with both t's and v's in the equation.

 

[haruspex]

I'm not familiar with this M, N, notation, so I cannot follow your method.
I would start by considering it dimensionally. If v is dimensionally some power n of t, what value of n makes the equation dimensionally consistent? That should suggest a change of variable that simplifies it a bit.
See how you go from there.

 

[Ssnow]

The equation $2t^2v'-2tv=-v^3$ can be written as $v'-\frac{v}{t}=-\frac{v^3}{2t^2}$ , that is a Bernoulli equation of type:

$y'(x) + f(x)y=g(x)y^{n}$

here $n=3, x=t, y=v, f=-\frac{1}{t}, g=-\frac{1}{2t^2}$. To solve it see the link

https://en.wikipedia.org/wiki/Bernoulli_differential_equation

 

[HallsofIvy]

The equation 2t^2v'= 2tv- v^3 can e written 2t^2 dv= (2tv- v^3)dt and then 2t^2dv+ (v^3- 2tv)dt. Many textbooks write the "generic" first order differential equation M(x,y)dy+ N(x,y) dx= 0 or, with variables v and t rather than y and x, M(v,t)dv+ N(v,t)dt.
Further, such an equation is "exact" (and so particularly easy to solve) if and only if \frac{\partial M}{\partial x}= \frac{\partial N}{\partial y} or, in terms of v and t, \frac{\partial M}{\partial t}= \frac{\partial N}{\partial v}. That is where Joseph1739 got "M= 2t^2" and "N= 2tv- v^3. Of course, as he pointed out, \frac{\partial M}{\partial t}= 4t\ne 2t- 3v^2= \frac{\partial N}{\partial v} so this is not an exact equation.