Recent content by Joseph1739

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'm confused by your hint. Didn't you just prove that it approaches 1?

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...

(a1)(1-rn/1-r) = (1)(1-xk/1-x) = (1-xk)/(1-x)

That makes sense, but how does 2a-1 factor out of (2a)b-1? 2 raised to a power minus 1 will not be zero anymore unless the exponent is 0.

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

Would't that result in = 1(1-xk)/(1-x) = (1-x)k-1?

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.

I was trying it out with (22)3 and it simplifies to the same result. I just really wanted to know how that equivalence was derived.