Recent content by cloud360

Fy=-2(t^2-2)y+3y^2-1 i have no idea how to show this is continuous? is there a method. if so i don't know what it is, cany you give me a link to a website which possibly teaches you how to know if somethig is continous. i know how to prove a function is 1-1 though

I did differentaite w.r.t y, see wolfram http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29 then i sub t=1 y=2 and got 15

First y’’=-2(t^2-2)y+3y^2-1, http://www.wolframalpha.com/input/?i=-%28t^2-+y-+2%29%28y-+1%29%28y%2B+1%29 at t=1 and y=2, we get that the only solution is 15. So the point (1,2) gives a unique solution, as required...

ok, i am very grateful for your help. i have my exam in 3 days. can you please kindly tell me how do i answer part e. wether there is a unique solution through the point (1,2). i know that to prove uniqueness, we must have y'=f(t,y), and differentiate w.r.t to y. i.e find (f(t,y))' and...

no. but i guess i have to explain that in the exam. why is it? i just thinking of positive as moving forwards, thus ==> and negative as moving backwards, thus <==

y'=-(t^2- y- 2)(y- 1)(y+ 1) -1<y<1 for large t, y' is positive y<-1 for large t,y' is negative y>1 for large y, y' is negative <==-1==>1<== so y=1 is a stable, y=-1 is unstable but isn't this just the same as my method in post 3, except a small error. all i have learned here is that lower...

but if i differentiate w.r.t t i get 0 http://www.wolframalpha.com/input/?i=differentiate+with+respect+to+t+dy%2Fdt+%3D+y2+-+1 this is so confusing. because i have not seen this kind of differentiation. the answer is definitely 0. if it isnt, then i have no idea how to get the solution i put...

thanks again for response. when differentiating. do i have to do implicit differentiation? i differentiated with respect to t, because you said "I have no idea how you got that for y''. One apparent mistake is that you differentiated with respect to y, but you have to differentiate with...

also, for the last question on this page http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf it says to determine wether the fixed points of y'=y^2-1 is stable or not where y'=y dot=dy/dt but if we differentiate again we get y''=0? does this mean all...

ok, for the question in the opening. the one about stability. if i differentiate with respect to t. i get -2ty^2+2t. and t=1 gives -2y^2+2=0 and y is imaginery. does this mean at t=1, point is unstable?also at t=-1, i get 2y^2-2==>y^2=1 i got y=+/- 1?? how do i know if it is stable or not?

furthermore, when y'=y^2-1,this does not work as y''=2y we have fixed points -1 and 1 V''(-1)=-2<0, so unstable but -1 is actually stable. according to website below. see last question. http://www.maths.manchester.ac.uk/~ahazel/MATH10232/Coursework_Test0910_Sol.pdf but my method of...

yes sorry V''(x)>0 is stable , as it is a minimum, but still i don't think this has got anything to do with this question. because the other 1 was related to potential energy. or am i wrong? calculating y'' gives -2y+3y^2+4y. this equation contains all y's? not x's? so how can we sub x, it...

you said (1-y)^2=(y-1)(y+1) but that is false (1-y)^2=-(y-1)(y+1)=(1-y)(y+1) see here: http://www.wolframalpha.com/input/?i=%281-y^2%29 so doesn't that mean your solution is wrong?

actually, the other time i was dealing with potential energy, and the notes said V(x)>0 is stable (minimum) where V=potential energy equation

ok i followed what your saying. can you tell me if my solution below is correct? for part c and d i said C is stable, and so is D