Math Help - differencial equation

1. differencial equation

sd

2. It's a separable ODE.

Try it.

3. 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?!

4. 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

5. The RHS is

how did you get the last part with 1-3/y+3?

6. $\frac{y}
{{y + 3}} = \frac{{y + 3 - 3}}
{{y + 3}} = 1 - \frac{3}
{{y + 3}}$

7. 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?

9. 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

10. 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?

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

look similar to the x integral?

i dont understand what you mean?

11. $\int {\frac{1}
{{y + 3}}~dy} = \int {\frac{{(y + 3)'}}
{{y + 3}}~dy} = \ln \left| {y + 3} \right| + k~\blacksquare$

Girl, it's quite dangerous getting in ODE if you don't remember how to integrate.

12. [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

13. 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

14. 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?

15. Originally Posted by emily28
sd
You know you shouldn't do that?