# powers

• Feb 13th 2008, 04:31 AM
dankelly07
powers
Very quickly, if your taking powers across the other side eg deifferential equations, how does it work again, im having a blank!!!

eg
solving for y

-y^-5 = 6x+C

becomes

y = (6x+C)-1/5

IS this right??

If it is can anyone explain the thoery behind it???
• Feb 13th 2008, 04:37 AM
dankelly07
ah man i think i wrote the question out wrong...

can anyone help me find the solution for the separable equation
6y^6

for y(0)=3
(1/243 - 30x)^-(1/5)

but can anyone help me through the stages?
I just need explicit help with this solution, I keep going wrong with the basic Algebra i think
• Feb 13th 2008, 04:41 AM
vperera
It should be y=(6x+C)^-(1/5).

Just multiply the exponent by its inverse (like if the exponent is 2, multiply both sides of the equation by 1/2).
• Feb 13th 2008, 04:46 AM
topsquark
Quote:

Originally Posted by dankelly07
can anyone help me find the solution for the separable equation
6y^6

This is not an equation. What is your original equation?

-Dan
• Feb 13th 2008, 02:21 PM
dankelly07

The equation I asked to solve was

dy/dx = 6y^6

find the general solution
w/ condition y(0) = 3
• Feb 13th 2008, 03:19 PM
mr fantastic
Quote:

Originally Posted by dankelly07

The equation I asked to solve was

dy/dx = 6y^6

find the general solution
w/ condition y(0) = 3

$\displaystyle \frac{dx}{dy} = \frac{1}{6} \frac{1}{y^6} = \frac{1}{6} y^{-6}$.

Therefore:

$\displaystyle x = \frac{1}{6} \left( \frac{-1}{5} \right) y^{-5} + C$

$\displaystyle = -\frac{1}{30} \frac{1}{y^5} + C$

$\displaystyle \Rightarrow 30(x - C) = \frac{1}{y^5}$.

You know that when x = 0, y = 3:

$\displaystyle -30C = \frac{1}{3^5} = \frac{1}{243}$.

Therefore $\displaystyle 30x - \frac{1}{243} = \frac{1}{y^5}$.

You can finish it off if you want to make y the subject .....
• Feb 14th 2008, 06:20 AM
dankelly07
Yeah thats gr8, I cant see where I was going wrong, I'm v bad at factorising!