If dy/dx=y/3*x^(1/3) and y=1 when x=1 then y=?

Question

The answers are as follows...
A) .5x^(2/3)-1
B) .5x^(2/3)-.5
C) e^(.5x^(2/3)-1)
D) e^(.5x^(2/3)-.5)
E) None of these
It would be great if someone could show me step by step how to solve this problem...Thanks

Answer 1

First, you need to seperate the variables, so:
dy/dx=y/3*x^(1/3) becomes dy/(y/3) =x^(1/3)dx
So, integrating both sides gives 3 ln y = (3/4) x^(4/3) +C. Now, y =1 when x=1,
so 3 ln 1 = 3/4 +C
Since ln 1=0, 0=3/4+C So, C= -3/4
Therefore: 3 ln y = (3/4) x^(4/3) -3/4
rewrite (3 ln y) as ln (y^3), then take
e^[ln (y^3)] = e^[(3/4) x^(4/3)-3/4].
So y^3 = e^[(3/4) x^(4/3)-3/4]
Finally, taking the cube root of both sides gives a solution for y.

Answer 2

This is a variables separable type of differential equation. You can use basic algebra to collect everything involving y (including the dy) on one side and everything involving x (including dx) on the other side. You then integrate each side separately (not forgetting to include a constant of integration on one side). You should then be able to rearrange the result into y = f(x) (although this is not always possible). The condition x = 1, y = 1 will allow you to work out the value of the constant of integration in this particular case.

Answer 3

I thnk your question is y' = y/ (3 * x^(1/3) )
(you do have to use brackets when you cant typeset)

y'/y = (1/3) x ^ (-1/3)
ln y = (1/3) x^(2/3)/(2/3)+ c =(1/2) x^(2/3) + c
y=1 when x=1 so 0 = 1/2 + c
ln y = (1/2) [ x^(2/3) - 1]
y = exp [ 0.5 (x^(2/3) -1) ]
so C