argh i dont understand!?

Question

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

Answer 1

You need to integrate that formula. First of all, expand it:
(3-√x)²/√x
(9 - 6√x + x)/√x
9/√x - 6 + x/√x
9x^-0.5 - 6 + x^0.5
Now integrate:
y = 18x^0.5 - 6x + 2/3 x^1.5 + c
Substitute in the point:
2/3 = 18 - 6 + 2/3 + c
c = -12
So y = 18x^0.5 - 6x + 2/3 x^1.5 - 12.

Answer 2

dy/dx = (3-√x)²/√x ,x>0

dy/dx = (9-6√x+x)/√x = 9/√x - 6 + √x

integrate that and you should get something like:
y= 18√x -6x +(2/3) x^(1.5) + c

plug in the values y=2/3 at x=1:
2/3 = 18 - 6 + 2/3 + c
c = -12

so y= 18√x -6x +(2/3) x^(1.5) -12

Answer 3

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 4

i think u need 2 distrubte after substuting the variables but wat do i no im only 13