- #1
Joseph1739
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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.
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.