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Use

1. v = y'

2. v' = y''

3. dy/dt = v

to solve the differential equation.

Question: 2t^{2}y'' + (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:

2t^{2}v' + (v^3) - 2tv = 0

M(t,v) = v^{3}- 2tv

N(t,v) = 2t^{2}

M_{v}= 3v^{2}- 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.

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# Equations with dependent variable missing

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