Recent content by 0131413

Tyvm. After reading the chapter I was so confused at first, but with your words I had a lightbulb-goes-on-moment and now everything makes sense. Wish I could give you a hug. :!)

Is the relation abs(x) = abs(y)? ...Are the equivalence classes to look like [x] = {-x, x}? :rolleyes:

In this case is there one equivalence class (depending on how the parabola is shifted) with an infinite amount of elements? From the way Wiki shows rational numbers, I played around with the parabola coordinates and is the relation something like y0-x02=y1-x12=y2-x22=...=yn-xn2? ...If the above...

Homework Statement Let f: R -> R, x -> x^2 What does the partition for the equivalence relation of this function look like? Homework Equations The Attempt at a Solution Uh...I have no idea. Sorry, the book only has examples of like integers from modulo n, if anybody could...

http://hps.org/publicinformation/ate/q7878.html Apparently you need the penetration of the coring device...convenient.

Thanks a lot. I understand now. I was not sure about the fundamental set when dealing with tan(x), but I see it now. SVXX: Because the lesson was on variation of parameters. :wink:

Homework Statement A particular solution of y'' + 4y = tanx Answer choices are: (a) 1/2*cos(2x)ln|sec(2x)+tan(2x)| (b) -1/2*cos(2x)ln|sec(2x)+tan(2x)| (c) 1/2*sin(2x)(ln*cos(x)+x*sec(2x)) (d) 1/2*sin(2x)(ln*cos(x)-x*sec(2x)) (e) none of the above Homework Equations The...

C2=-3 The equation that satisfies the problem is y=-xe^(-x)-3e^(-x) y'=xe^(-x)-e^(-x)+3e^(-x) y''=-xe^(-x)+e^(-x)+e^(-x)-3e^(-x) -xe^(-x)+2e^(-x)-3e^(-x)+xe^(-x)-e^(-x)+3e^(-x)=e^(-x) Final answer to the question of what value of a satisfies the problem: a=-2 Thanks a lot. :)

I'm sorry that t is supposed to be an x. I did forget the negative sign when trying to solve the general solution, ouch. I thought it would be xe^(-x) since using yp=Ae^(-x) I ended up with Ae^(-x)-Ae^(-x). Now I have y=-xe^(-x)+C1+C2e^(-x) y'=xe^(-x)-e^(-x)-C2e^(-x) y(0)=C1+C2=1...

Homework Statement What is the value of a such that the solution of the initial-value problem satisfies limx->infinity y(x) = 0? y''+y'=e^(-x), y(0)=1, y'(0)=a Homework Equations The Attempt at a Solution Not sure what to do with the missing y term... yp=Ae^(-x), y'p=-A^(-x), y''p=A^(-x)...