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First Order ODE With v = y' Substitution

  1. Nov 15, 2014 #1
    1. The problem statement, all variables and given/known data

    8t^2 * y'' + (y')^3 = 8ty' , t > 0

    2. Relevant equations


    3. The attempt at a solution

    I tried using the substitution v = y' to get:

    8t^2 * v' + v^3 = 8tv

    I rewrote it in the form 8t^2 * dv/dt + v^3 = 8tv, and then moved the v^3 to the other side to get 8t^2 * dv/dt = 8tv - v^3.

    I then multiplied both sides by dt to get:

    8t^2 dv = (8tv - v^3) dt

    Try to get it in exact form.

    8t^2 dv - (8tv - v^3) dt = 0

    8t^2 dv + (8tv + v^3) dt = 0

    But when I test for exactness, it doesn't work, because 16t doesn't equal 3v^2 + 8t. Did I mess up somewhere, or is there some way to continue?
     
  2. jcsd
  3. Nov 15, 2014 #2

    HallsofIvy

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    First write the equation as [itex]8t^2dv+ (v^3- 8tv)dt= 0[/itex].
    Yes, that is not an "exact equation" because the derivative of [itex]8t^2[/itex], with respect to t is [itex]16t[/itex] while the derivative of [itex]v^3- 8tv[/itex] with respect to v is [itex]3v^2- 8t[/itex]. They are not the same. However, since everything here is a power of t and v, I would try powers of t and v as an integrating factor- multiply the both sides of the equation by [itex]t^nv^m[/itex] to get [itex]8t^{n+2}v^mdv+ (t^nv^{m+3}- 8t^{n+1}v^m)dt= 0[/itex]. Do there exist values of m and n such that this is an exact equation?
     
  4. Nov 15, 2014 #3

    LCKurtz

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    Have you studied the Bernoulli equation? If so, if you write it as ##8t^2v' - 8tv = -v^3## you will recognize that dividing by ##v^3## and letting ##u = v^{-2}## will lead you to a linear first order DE.
     
  5. Nov 15, 2014 #4
    Alright, let's see.

    8t^2v' - 8tv = -v^3

    v' - v*t^-1 = -v^3 / (8t^2)

    v^-3 * v' - v^-2 * t^-1 = -1 / (8t^2)

    Bernoulli form is v^-n * v' + p(t)*v^(1-n) = q(t)

    So that means n = 3 in this case.

    Use a substitution u = v^(1 - 3)

    u = v^-2 and u' = -2v^-3*v'

    So then substituting back in, we get:

    -0.5u' - u/t = -1/(8t^2)

    So then I multiply both sides by -2 and get:

    u' + 2u/t = 2/(8t^2)

    u' + 2u/t = 1/(4t^2)

    Integrating factor is e^(∫(2/t)dt)

    Integrating factor is e^(2*ln(t))

    e^(ln(t^2)), which becomes t^2

    Multiply both sides by the integrating factor.

    t^2 * u' + 2tu = 1/4

    So t^2 * u = ∫(t^2*0.25t^-2)

    t^2*u = ∫0.25 dt

    t^2*u = 0.25t

    u = 0.25t / (t^2)

    u = 0.25/t + c

    u = v^-2, so v^-2 = 0.25/t + c

    Flip both sides.

    v^2 = 1/(0.25/t + c)

    v^2 = (t / c1t + 0.25)

    v = √(t / c1t + 0.25)

    Since v = y', that means y = ∫√(t / c1t + 0.25). Am I on the right track once I solve that integral?
     
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