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January 30, 2015

January 30, 2015

Posted by **ANONYMOUS** on Wednesday, May 30, 2012 at 8:57am.

I can't figure out what to do with that cubed root; it keeps throwing me off.

- CALCULUS -
**Damon**, Wednesday, May 30, 2012 at 9:19amdy/dx = x y^(1/3)

y^(-1/3) dy = x dx

(3/2)y^(2/3) = (1/2)x^2 + c

when x = 2 , y = 8

(3/2)(4) = (1/2)(4) + c

c = 6 - 2 = 4

(3/2) y^(2/3) = (1/2) x^2 + 4

3 y^(2/3) = x^2 + 8

y^(2/3) = (1/3) x^2 + 8/3

- CALCULUS -
**ANONYMOUS**, Wednesday, May 30, 2012 at 9:38amAnd then you could simplify it out to get y = (x^2/3 + 8/3)^(3/2). Thank you so much, I went back and ran (2, 8) through and it finally worked out, thanks!!!

- CALCULUS -
**Damon**, Wednesday, May 30, 2012 at 9:52amyour wrong

- CALCULUS -
**Damon**, Wednesday, May 30, 2012 at 10:13amStrange, I did not write that.

- CALCULUS -
**ANONYMOUS**, Wednesday, May 30, 2012 at 10:39amAll I did was solve for y so I could get the y by itself. I need to right the expression in terms of x, so I need to isolate y completely. In order to do this, I just applied the 2/3 root to both sides of the equation. Put in different terms, I could also say that it is to the 3/2 power. I worked it through with the initial condition f(2)=8, and it worked out. So, would you put your stamp of approval on this simplification.

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