argh i dont understand!?

Question

it says "given that dy/dx = (3-√x)²/√x ,x>0, and that y=2/3 at x=1, find y in terms of x" i really don't understand!

Answer 1

dy/dx=(3-√x)²/√x
=(9+x-6*(x)^(1/2)/x^(1/2)
=(9x^(-1/2)+x^(1/2)-6)
dy=(9x^(-1/2)+x^(1/2)-6)dx
integrate
y=int{9x^(-1/2)+x^(1/2)-6}dx
= 18x^(1/2)+2/3x^(1.5)-6x+D
use initial conditions
2/3=18+2/3-6+D
>>D= -12

therefore,
y=2/3x^(1.5)-6x+18x^(1/2)-12

i hope that this helps

Answer 2

I think it would be best to expand your dy/dx.

dy/dx = (9 - 6√x + x) / √x = 9x^(-1/2) - 6 + x^(1/2)

y = 9x^(1/2)/(1/2) - 6x + x^(3/2)/(3/2) + C
y = 18x^(1/2) - 6x + (2/3)x^(3/2) + C

Substitute x=1 and y=2/3,
2/3 = 18 - 6 + (2/3) + C
C = -18 + 6
C = -12

So,
y = 18x^(1/2) - 6x + (2/3)x^(3/2) - 12
y = 18√x - 6x + (2/3)(√x)^3 - 12

Answer 3

what it is asking is that you rearrange the formula so that y is the subject and then work out its value. Remember, if you transfer something to the other side of the equals sign you change its sign to the opposite. Alternativly you could put in the values and work it out from there.
I gopt that :
y=9+x -6√x
and then you have to work out the bit about the other stuff, after that i get abit confused too and can't help you.

Answer 4

No, neither do I.

Answer 5

???? that is just crazy

Answer 6

hmmm...

Answer 7

i don't know!!!