1. separable equation

Find particular solution of the separable equation

$6(x^4-3x)dx+\frac{1}{y^3}dy=0$

2. ok, so now I have some idea how this LATEX works, I will try to post my attempted solution

3. Sorry,it's obvious that I haven't got to grips with Latex

4. $(6x^4-18x)dx=(-y^{-3})dy$

$\int(6x^4-18x)dx=-\int(y^{-3})dy$

$\frac{6}{5}x^5-9x^2+c=-\frac{-y^-2}{-2}$

$\frac{6}{5}x^5-9x^2+c=\frac{1}{2y^2}$

Looks right except for that you need a 2 on the final step

5. I forgot to say that at the beginning I am told that y(0)=1

I will try to show my solution later.Sorry for being a nuisance.

6. $(6x^4-18x)dx=(-y^-3)dy$

$\int(6x^4-18x)dx=\int(-y^-3)dy$

$\frac{6}{5}x^5-9x^2+c=\frac{1}{2y^2}$

$y(0)=1$

$\frac{6}{5}(0)^5-9(0)^2+c=\frac{1}{2(1)^2}$

$c=\frac{1}{2}$

Now I have:

$\frac{6}{5}x^5-9x^2+\frac{1}{2}=\frac{1}{2y^2}$

7. Should I stop there, or proceed to express it as y=

like this:
multiplying across by 10 , I get:

$12x^5-90x^2+5=\frac{5}{y^2}$

Therfore:

$y^2(12x^5-90x^2+5)=5$

$y=\frac{\sqrt5}{\sqrt(12x^5-90x^2+5)}$

8. looks good.