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Equations with dependent variable missing

  1. Feb 9, 2016 #1
    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
    Mv = 3v2 - 2t
    Nt = 4t
    Since these two are not equal, I have to find an integrating factor, but using both:
    (1) dμ/dt = [(Mv-Nt)/(N)] * μ
    (2) dμ/dv = [(Nt-Mv)/(M)] * μ
    both result in a unsolvable equation because I end up with both t's and v's in the equation.
     
  2. jcsd
  3. Feb 10, 2016 #2

    haruspex

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    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.
     
  4. Feb 14, 2016 #3

    Ssnow

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    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
     
  5. Feb 15, 2016 #4

    HallsofIvy

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