Little of maths help please

[anksta]

Sposed to be revisin the third maths module for my resit in a couple of weeks but I swear these notes must be wrong:



can anyone confirm if thats right?

I can't see where the 2 goes and also why you come out with dx/x

cheers

 

[mrmoose]

the dx/x bit is just the x dx/dv = v+1 rearranged. as far as the other stuff goes, i can't renember, thank god uni is easier than this

 

[azteched]

It's correct. The 2v goes because you had v on the other side (2-1=1 ).
You get dx/x by multiplying through by the differential dx, and dividing by x.

 

[anksta]

the dx/x bit is just the x dx/dv = v+1 rearranged. as far as the other stuff goes, i can't renember, thank god uni is easier than this

yeah but if you just take the dv bit across doesnt that leave you with x dx = dv/(v+1)? And where has the 2 gone?

Also I'm goin into 3rd year at uni

 

[anksta]

It's correct. The 2v goes because you had v on the other side (2-1=1 ).
You get dx/x by multiplying through by the differential dx, and dividing by x.

I dont get where the v comes from to take it away though, its multiplyin on one side so I dont get why you can just subtract it on the other.

 

[azteched]

You have

dy/dx = 2v + 1 (first line)

and

dy/dx = v + x*dv/dx (second line, from differentiating y=vx wrt x).

Substitute the second into the first:

v + x*dv/dx = 2v + 1
x*dv/dx = v + 1

then it's just a case of multiplying/dividing by various factors.

 

[Arcade Fire]

I dont get where the v comes from to take it away though, its multiplyin on one side so I dont get why you can just subtract it on the other.

Notice that:

d[y/x]/dx = 1/x * dy/dx - 1/x * [y/x]

and then rearrange to get

dy/dx = x * d[y/x]/dx + [y/x]

and stick that in the equation you start with, to get

x * d[y/x]/dx + [y/x] = 2*[y/x] + 1

Now notice that one of the [y/x] terms cancels on each side. That's what happens to the 2.