Recent content by 5hassay

awesome. Also, I also think you are correct about the missing C, thank you--I shall update that post. Thank you for your help!

true. OK, then \displaystyle{\frac{dv}{dx} = \frac{v \log(v) - v}{x} \Longrightarrow \int{\frac{1}{v} \cdot \frac{1}{\log(v) - 1}} dv = \int{\frac{1}{x}}} dx \displaystyle{\Longrightarrow \log(\log(v) - 1) = \log(x) + C \Longrightarrow \log(v) - 1 = C x \Longrightarrow v = \frac{y^{\prime}}{x}...

thanks for the help. I don't know how to evaluate \frac{d \log(u)}{d \log(x)} I understand that I would be differentiating \log(u) as a function of \log(x), but I don't see how that is a function of \log(x).

thanks for the help. OK, that was productive. Here's what I did: v := \frac{y^{\prime}}{x} \Longrightarrow v^{\prime} = \frac{x y^{\prime \prime} - y^{\prime}}{x^2} \Longrightarrow x^2 v^{\prime} + y^{\prime} = x y^{\prime \prime} Substituting, x^2 v^{\prime} + y^{\prime} =...

EDIT: my problem is solved, thank you to those who helped Homework Statement Solve: x y^{\prime \prime} = y^{\prime} \log (\frac{y^{\prime}}{x}) Note: This is the first part of an undergraduate applications course in differential equations. We were taught to solve second order...

Gotcha; that makes a bit more sense now. Thanks for the generalization!

Thanks for the reply. Alright... I think I need to work on my intuition for this stuff to better understand your help. Anyway, I think I've got something decent, and I'm done with it. For anyone looking at this thread for help, I found this link helpful...

Thanks for the response. "If you sketch your curve and thing of things flowing left to right, they would be flowing in the direction of increasing r from your curve." I don't understand what you are saying here. Could you please explain more explicitly? I am thinking you are talking about...

Homework Statement Compute the flux of \overrightarrow{F}(x,y) = (-y,x) from left to right across the curve that is the image of the path \overrightarrow{\gamma} : [0, \pi /2] \rightarrow \mathbb{R}^2, t \mapsto (t\cos(t), t\sin(t)). A (2-space) graph was actually given, and the problem...

Assuming we are dealing in the realm of undergraduate linear algebra... Continuing on what WannabeNewton said, think about why \mathbb{R} is a subspace of itself, and then consider some nonempty set that is not \mathbb{R} or the set consisting of just 0. Recall that a nonempty set is a...

The horizontal and vertical line test is an intuitive way to see if a graph belongs to an injective function. To be clear, if the graph belongs to a function, and a horizontal line drawn anywhere on the graph means that the line will only intersect the graph at most once, then the function is...

thanks for the reply. I still find the concept odd, but it is a bit more clear, thanks. I think I just need time for this to settle in my mind, maybe. A few questions, though. One, when we are referring to the eigenvectors as e_1 and e_2, are we referring to them as [1,0] and [0,1], the...

Do you mean letting y=5? If so, then yes, you get \pm \sqrt{-1}, which is not a real number, and therefore there does not exist any x in the domain of f such that f(x)=5. Consequently, the range of f does not include 5, for example. That is enough to show that surjectivity is not held by f, as...

What is your reasoning? Substitute some values for x into f and see what you get, or what you can't get. Does your answer change? Even better, use graphing software to visualize f, and then everything should be pretty clear. (Example, search "wolframalpha" into a search engine and type "f(x) = 4...

Your notation for the function definition isn't correct, I think. I believe you meant f : \mathbb{R} \rightarrow \mathbb{R}, x \in \mathbb{R} \mapsto 4 - x^2. Here, \mathbb{R} is both the domain and codomain of f. Surjectivity is the property that the image of the domain of f, which is...