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...
(df/dx) + (df/dy)* (dy/dx) = df(x,y)/dx
My book mentions the chain rule to obtain the right side of the equation, but I don't see how. The chain rule has no mention of addition. The furthest I got was applying the chain rule to the right operant resulting in:
df/dx + df/dx = 2(df/dx)
I still don't understand what I'm looking for. What was the purpose of taking the log of the expression? I'm guessing that constant is the base of the logarithm? I don't understand where that comes from though.
I don't understand why the limit of x1/loga(x) as x approaches infinity is a, where a can be any constant for the base.
Why isn't it 1? The base (x), approaches infinity, while the exponent approaches 0 (1/infinity), so it should be (infinity)0 = 1.
Suppose I'm at this point in solving a differential equation and the initial condition is Q(0) = Q0
-ln|25-Q| + c1 = rt/100 + c2
Then if I combine c2-c1, I can rename it to c, we have:
-ln|25-Q| = rt/100 + c
Now if I multiply the equation by (-1), I get:
ln|25-Q| = -rt/100 - c
If I let -c = C...
That makes sense. Is the chain rule even applicable here though? dy/dt is not a fraction, so I don't see how you can multiply by dt then cancel the dt's. Then I don't believe dt is a derivative (isn't it a small change in t?), so how can the chain rule be used here? I don't want to struggle to...
Not sure if this is the correct place to post this.
dy/dt = 0, find y(t)
My professor told me that the chain rule is used to determine that (dy/dt)*dt = dy, but I just don't see it.
Multiply both sides by dt.
(dy/dt)*dt = 0dt
(dy/dt)*dt = 0
dy = 0, then integrating both sides:
y = C
dy/dt is...
Yeah, I did google geometric series. I looked at the sum of geometric series formula when the other guy posted about it. And I got:
1 = first element
x = common ratio
k = nth element
I don't know how that expansion is derived. I understand how to get expression on the right side by dividing the original polynomial by (x-1), but I don't know how (x-1) was factored with begin with.
Would that make it: ((2ab(1-2ab)/(1-2)) -1 = -22ab -1 ? Am I doing something wrong here? Ignoring the -1 right now, the first term is 2ab, n = ab, and the common ratio is 2.