# separable equation

• Aug 10th 2010, 08:17 AM
redskies
separable 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 AM
redskies
ok, so now I have some idea how this LATEX works, I will try to post my attempted solution
• Aug 10th 2010, 08:35 AM
redskies
Sorry,it's obvious that I haven't got to grips with Latex :(
• Aug 10th 2010, 08:56 AM
11rdc11
$\displaystyle (6x^4-18x)dx=(-y^{-3})dy$

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

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

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

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

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

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

$\displaystyle y(0)=1$

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

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

Now I have:

$\displaystyle \frac{6}{5}x^5-9x^2+\frac{1}{2}=\frac{1}{2y^2}$
• Aug 10th 2010, 10:06 AM
redskies
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 AM
11rdc11
looks good.