Find particular solution of the separable equation

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

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- Aug 10th 2010, 08:17 AMredskiesseparable equation
Find particular solution of the separable equation

$\displaystyle 6(x^4-3x)dx+\frac{1}{y^3}dy=0$ - Aug 10th 2010, 08:19 AMredskies
ok, so now I have some idea how this LATEX works, I will try to post my attempted solution

- Aug 10th 2010, 08:35 AMredskies
Sorry,it's obvious that I haven't got to grips with Latex :(

- Aug 10th 2010, 08:56 AM11rdc11
$\displaystyle (6x^4-18x)dx=(-y^{-3})dy$

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

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

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

Looks right except for that you need a 2 on the final step - Aug 10th 2010, 08:58 AMredskies
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. - Aug 10th 2010, 09:57 AMredskies
$\displaystyle (6x^4-18x)dx=(-y^-3)dy$

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

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

$\displaystyle y(0)=1$

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

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

Now I have:

$\displaystyle \frac{6}{5}x^5-9x^2+\frac{1}{2}=\frac{1}{2y^2}$ - Aug 10th 2010, 10:06 AMredskies
Should I stop there, or proceed to express it as y=

like this:

multiplying across by 10 , I get:

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

Therfore:

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

$\displaystyle y=\frac{\sqrt5}{\sqrt(12x^5-90x^2+5)}$ - Aug 10th 2010, 10:06 AM11rdc11
looks good.