Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

thank you for the information...

god bless..

btw, do you have any reference where i can get a sumary of trig identity? thanks again...

phi as in 180 deg... or is it pi

**seiya_001**- Replies: 6

can anybody tell me how to change

1 phi

- ( ∫ A sin x . cos nx dx)

phi 0

into:

A 1 phi

- . - . ∫(sin (1+n)x + sin (1-n)x )dx

phi 2 0

thanks before...

**seiya_001**- Replies: 4

umm..can anybody tell me some line (or is it curve..) equations? you know stuff like y=2x is linear equation.. i know it's lame but i really need it.. thanks

thanks a lot dude..i'll try to find it in my country.. i'm sure we have it somewhere...:)

kybasche wrote:

I don't know if you have access to a Diff.Eq. book, but there should be a list of the different things that you can end up with depending on the right hand side.

any good reference? im' in indonesia so it's rather hard to get a good book with the original language (usually they're translated into indonesian and sometimes it's confusing), but shoot anyway...?

kybasche wrote:

Have you solved diffeq's before? If not, I can see how it would seem a bit daunting smile

yes, but not with high order equations and all (d^2q/dt^2 or higher)

a question, what if the right hand side of the equation is not constant, let say it's a function also then what would you do with the 'particular solution'? thanks before..(btw in what year are you in college?)

thanks a lot dude..i'll need some time to understand it all but i'll try...:)

kybasche wrote:

in my opinion, the easiest way to solve this would be to use laplace transforms. do you know how to use them?

umm..not yet.., what about the other method you mentioned? can it be solved by any diffferential equation? thanx..:)

**seiya_001**- Replies: 10

help....

how do you find the Q from:

d^2q dq

------ + 8 ----- + 25 q = 300

dt^2 dt

any help would be MOST appreciated...

I hope this helps...

your problem is:

∫2x^2/(2x^2+1) dx

divide 2x^2/(2x^2+1) to acquire 1-(1/2x^2+1)

then: ∫ 1-(1/2x^2+1) dx = ∫dx - ∫(1/2x^2+1) dx, yay..an easier integration...!!

= x-∫(1/2x^2+1) dx, and we know that ∫1/(u^2+a^2) dx = 1/a arc tg u/a

thus: x- (arc tg x√2) or x(1-arc tg √2) + C

guys please CMIIW...

boombastictiger wrote:

4. A curve has the following equation: y=x³ -12x+7

(a)

(i) Find the gradient function ,dy?dx, of the curve. (2 marks)(ii) Find the coordinates of the points on the curve at which the gradient is zero. (6 marks)

(iii) Describe the nature of the curves turning points. (4 marks)

(

Challenging....

if y=x^3-12x+7

then

i) dy/dx = 3x^2 - 12

ii)dy/dx=0, 0=3x^2-12, 3x^2=12

x=√(12/3)=√4=2

y=2^3 - 12.2 +7

= 8-24+7=-9

iii) i don't understand the question, sorry..

guys, please correct me if i'm wrong...

**seiya_001**- Replies: 7

hi guys, i just find out about this awesome forum.. I'm a freshmen in one of the colleges in Indonesia and i love to learn about math...

Pages: **1**