Recent content by Dr Game

its V = IR

then its got to be \frac{1}{y^2}Tan^-^1+C

im confused could the integrual be \frac{1}{2y^2}ln|\frac{1-y^2}{1+y^2}| + C ?

I take it you are or where a metallurgist? You have found some pretty impressive resources, usually it's hard to find such descriptive things on the Internet... it compares quite well with the books I have on it

2Cx = \frac {2Cx^2}{x} 2Cx = 2Cx is that possible?

Homework Statement Verify that the function is a solution of the DE Dy = \frac{2y}{x} , y = Cx^2 2. The attempt at a solution LHS = Dy = D(Cx^2) = 2 Cx RHS = \frac{2y}{x} then... I really don't know what to do from there... do I just simply things?

alright thought so, so I should get the same answer if C wasn't there, which is 0 25Ce^-^5^x+4(-5Ce^-^5^x)-5Ce^-^5^x = 0

It's been a while, but how would you go about deriving y = C e^-^5^x and plug it into this equation y'' + 4y' - 5y = 0 i can do this without C the answer would be 0, but what do I do with C there?

you know what... I thought it was the number that comes out after you derive it a few times.. not how many times you derive it i get it now, thanks

because I don't understand

the following is an attachment of my notes: I don't get why the order is 2 and not -3

I'm almost done mech eng in Ottawa.. there where never any girls in our year, but there is one below my year.. and she said she's doing it because where she works she's an electrical engineer, and they've started to need people on more of the mechanical side, so they're paying her to take the...

I was taking notes in class and the prof said that in the equation y^(double prime)-3y^(prime)+2y = 0 that 2 was the order.. is that due to the following? the first y = 0 when you take the derivative twice the second y = 0 when you derive it once and the first one = 2 when you derive it...

wow.. I see how to do the second one.. its almost like a whole new order of operations to me... you derive first, then square it... I'm very gracious for you spending your time and effort with me, thank you

why do you only take the derivative once... is not everything else squared too? you have to derive it twice don't you to get rid of \frac{d}{dx}