differencial equation

Show 40 post(s) from this thread on one page
Page 1 of 2 12 Last
• Jul 11th 2007, 07:04 AM
emily28
differencial equation
sd
• Jul 11th 2007, 07:06 AM
Krizalid
It's a separable ODE.

Try it.
• Jul 11th 2007, 07:11 AM
emily28
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 AM
topsquark
Quote:

Originally Posted by emily28
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?!:(

Actually it should be:
$\frac{yy^{\prime}}{2y + 6} = \frac{1}{x}$

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

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

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

Can you do it now?

-Dan
• Jul 11th 2007, 07:20 AM
emily28
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 AM
Krizalid
$\frac{y}
{{y + 3}} = \frac{{y + 3 - 3}}
{{y + 3}} = 1 - \frac{3}
{{y + 3}}$
• Jul 11th 2007, 07:25 AM
emily28
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 AM
Krizalid
• Jul 11th 2007, 07:33 AM
topsquark
Quote:

Originally Posted by emily28
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?

Let me give you one more hint. Look at the RHS:
$\int \frac{1}{x} dx$

I presume that this integral is giving you no problems.

Is there any substitution you can think of that will make
$\int \frac{1}{y + 3}dy$
look similar to the x integral?

-Dan
• Jul 11th 2007, 07:38 AM
emily28
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 AM
Krizalid
$\int {\frac{1}
{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}
{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare$

Girl, it's quite dangerous getting in ODE if you don't remember how to integrate.
• Jul 11th 2007, 07:56 AM
emily28
[quote=Krizalid;60174] $\int {\frac{1}
{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}
{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare$

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 AM
Krizalid
Quote:

Originally Posted by topsquark
$\frac{1}{2} \int \left (1 - \frac{3}{y + 3} \right ) dy$

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

I guess we're done Emily :)
• Jul 11th 2007, 08:05 AM
emily28
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 AM
Krizalid
Quote:

Originally Posted by emily28
sd

You know you shouldn't do that?