find the solution to the initial value problem.
dy/dx=x((x^2)+1)/9y^3 y(0)=-1/sqrt(3)

Question

find the solution to the initial value problem. 
dy/dx=x((x^2)+1)/9y^3  y(0)=-1/sqrt(3)

zzr0ck3r · Answer

can you get all the x's on one side and all the y's on one side?

zzr0ck3r · Answer

$$\frac{dy}{dx}=\frac{x(x^{2}+1)}{9y^{3}}\rightarrow 9y^{3}*dy=x(x^{2}+1)dx$$

zzr0ck3r · Answer

you with me?

anonymous · Answer

yea

zzr0ck3r · Answer

$$\int9y^{3}dy=\int x(x^{2}+1)dx$$

zzr0ck3r · Answer

so
$$\int 9y^{3}dy=\int (x^{3}+x)dx$$

zzr0ck3r · Answer

can you solve this?

anonymous · Answer

9/4y^(4)=(x^(4)/4)+(x^(2)/2)+c

zzr0ck3r · Answer

right now solve for y

anonymous · Answer

y=((x^(4)/9)+(2x^(2)/9)+(4/9c))^(1/4)

zzr0ck3r · Answer

now use the initial condition to solve for c

zzr0ck3r · Answer

so plug in 0 for x, and -1/sqrt(3) for y

anonymous · Answer

is c=-1/4?

zzr0ck3r · Answer

use 9/4y^(4)=(x^(4)/4)+(x^(2)/2)+c
plug in the point (0,-1/sqrt(3))
9/4(-1/sqrt(3))^(4)=(0^(4)/4)+(0^(2)/2)+c
c=1/4

zzr0ck3r · Answer

the negative goes away

anonymous · Answer

oh right. careless mistake

anonymous · Answer

is it ok if i send you two other problems as an attached file and can you help me look over them because i dont know where my error is?

zzr0ck3r · Answer

sure, do you know if we got this one right?

zzr0ck3r · Answer

I trusted your integration so it should be fine..

anonymous · Answer

im going to type in the answer right now give me a moment :)

anonymous · Answer

i got it wrong. i'll attach my work and show you.

zzr0ck3r · Answer

the one we just did?

anonymous · Answer

yes

zzr0ck3r · Answer

hmm let me write it on paper real fast, I hate using the latek here I am no good...

anonymous · Answer

attachment

anonymous · Answer

i'll be back in 5 minutes

zzr0ck3r · Answer

$$y=\sqrt[4]{\frac{x^{4}+2x^{2}+1}{9}}$$
is this what you got?

anonymous · Answer

i had a 4/9c at the end is that wrong? (i attached the file of my work)

zzr0ck3r · Answer

sorry on phone, give me a few minutes...my wife is lost lol

anonymous · Answer

yeah its fine

anonymous · Answer

wait nevermind. yes i got that answer you got at the end

zzr0ck3r · Answer

oh ok, good. :)

zzr0ck3r · Answer

need anything else?

anonymous · Answer

yes just one problem. i'll attach it.

anonymous · Answer

actually im still working on it right now. i actually got it though thank you for your help :)

zzr0ck3r · Answer

good deal
One thing to look out for. 
say we have 3y=3x+c
and we divide by 3
y=x+c/3
many people think (because we can do this with normal integration for a general solution), that we can just say y=x+c because after all c is a constant, don't do this. Keep c separated and then solve for c when you have a ivp. You didn't have this problem but it is something I just remembered.