separable equation

**redskies** · August 10th 2010, 09:17 AM

Find particular solution of the separable equation

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

**redskies** · August 10th 2010, 09:19 AM

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

**redskies** · August 10th 2010, 09:35 AM

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

**11rdc11** · August 10th 2010, 09:56 AM

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

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

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

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

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

**redskies** · August 10th 2010, 09:58 AM

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.

**redskies** · August 10th 2010, 10:57 AM

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

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

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

$y(0)=1$

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

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

Now I have:

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

**redskies** · August 10th 2010, 11:06 AM

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:

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

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

**11rdc11** · August 10th 2010, 11:06 AM

looks good.

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