sd

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- Jul 11th 2007, 07:04 AMemily28differencial equation
sd

- Jul 11th 2007, 07:06 AMKrizalid
It's a separable ODE.

Try it. - Jul 11th 2007, 07:11 AMemily28
ive been trying it for nearly an hour, but can only get to the stage of

yy'/2y+3=1/x

because i havent dont much of these type of maths. where should i go now?!:( - Jul 11th 2007, 07:16 AMtopsquark
Actually it should be:

$\displaystyle \frac{yy^{\prime}}{2y + 6} = \frac{1}{x}$

$\displaystyle \frac{y}{2y + 6} dy = \frac{1}{x}dx$

$\displaystyle \int \frac{y}{2y + 6} dy = \int \frac{1}{x}dx$

The RHS is

$\displaystyle \int \frac{y}{2y + 6} dy = \frac{1}{2} \int \frac{y}{y + 3} dy = \frac{1}{2} \int \left (1 - \frac{3}{y + 3} \right ) dy $

Can you do it now?

-Dan - Jul 11th 2007, 07:20 AMemily28
The RHS is

http://www.mathhelpforum.com/math-he...3a0fb749-1.gif

how did you get the last part with 1-3/y+3? - Jul 11th 2007, 07:22 AMKrizalid
$\displaystyle \frac{y}

{{y + 3}} = \frac{{y + 3 - 3}}

{{y + 3}} = 1 - \frac{3}

{{y + 3}}$ - Jul 11th 2007, 07:25 AMemily28
oh right, this is slightly embarrasing becuase you are all so good at maths, but i still not sure where to go next. i should work on the LHS right?

- Jul 11th 2007, 07:29 AMKrizalid
Show your progress :D:D

- Jul 11th 2007, 07:33 AMtopsquark
Let me give you one more hint. Look at the RHS:

$\displaystyle \int \frac{1}{x} dx$

I presume that this integral is giving you no problems.

Is there any substitution you can think of that will make

$\displaystyle \int \frac{1}{y + 3}dy$

look similar to the x integral?

-Dan - Jul 11th 2007, 07:38 AMemily28
well i know the intergral of 1/x is |ln|x+c so thats the right side sorted:) so im making progress!! (not much but hey its better than nothing right?:rolleyes:

as for the LHS Is there any substitution you can think of that will make

http://www.mathhelpforum.com/math-he...53cf712d-1.gif

look similar to the x integral?

i dont understand what you mean? - Jul 11th 2007, 07:48 AMKrizalid
$\displaystyle \int {\frac{1}

{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}

{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare$

Girl, it's quite dangerous getting in ODE if you don't remember how to integrate. - Jul 11th 2007, 07:56 AMemily28
[quote=Krizalid;60174]$\displaystyle \int {\frac{1}

{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}

{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare$

i understand how to do this, this is simple intergration but dont understand how to relate it to 1/2 intergral (1-3/(y+3) dy - Jul 11th 2007, 08:00 AMKrizalid
- Jul 11th 2007, 08:05 AMemily28
sorry intergration has never been my strong point! hence all the questions :P

so if i put it all together i get

1/2 (y-3ln|y+3|)=ln|x|+c

so (y-3ln|y+3|)=2ln|x|+2c is my final answer? - Jul 20th 2007, 07:00 AMKrizalid