need helping solving diff eq.
dy/dx + y^1/3 = 0 y(0) = Yo

Question

need helping solving diff eq.
dy/dx + y^1/3 = 0        y(0) = Yo

ganeshie8 · Answer

separate variables ?

anonymous · Answer

So I subtracted y^1/3 from both sides then seperated..

anonymous · Answer

dy/y^1/3 = -1dx

ganeshie8 · Answer

looks good, integrate

anonymous · Answer

this is where I am having some problems.......do I just move y up top as y^-1/3....this is where I am differeing from my classmates.

ganeshie8 · Answer

$$\frac{dy}{y^\frac{1}{3}} =  -1 dx$$

is same as

$$y^\frac{-1}{3} dy =  -1 dx$$

anonymous · Answer

so the integral on left side would be -1/3*y^2/3 right? My classmates are smehow getting 3/2y^2/3

ganeshie8 · Answer

$$\large  \int  y^\frac{-1}{3} dy =  \int  -1 dx$$

$$\large  \dfrac{y^{\frac{-1}{3}+1} }{\frac{-1}{3}+1} = -x + c$$

$$\large  \dfrac{y^{\frac{2}{3}} }{\frac{2}{3}} = -x + c$$

$$\large  \frac{3}{2}y^{\frac{2}{3}}= -x + c$$

anonymous · Answer

Thanks.....its been a long day...obviously lol

ganeshie8 · Answer

:) you still need to plugin the initial value and solve $c$

anonymous · Answer

i got it (;