Another differential equation

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October 26th 2008, 06:40 AM
JrShohin

Another differential equation

Hi,

I'm solving this exercise, but i can not continue it, please, could you show me what to do with log.
thank you in advance.

http://i045.radikal.ru/0810/71/7f506ae6069d.jpg
y(0) = 0
October 26th 2008, 06:48 AM
Chris L T521

Quote:

Originally Posted by JrShohin

Hi,

I'm solving this exercise, but i can not continue it, please, could you show me what to do with log.
thank you in advance.

http://i045.radikal.ru/0810/71/7f506ae6069d.jpg
y(0) = 0

I would leave it as $\frac{1}{4}\log\left|4y+1\right|=\frac{x^2}{2}+C$

Now, multiply both sides by 4 and then get rid of the log:

$\log|4y+1|=2x^2+C\implies e^{\log|4y+1|}=e^{2x^2+C}\implies 4y+1=Ce^{2x^2}$

Now solve for y. Then apply the initial condition.

Can you take it from here?

--Chris
October 26th 2008, 07:13 AM
JrShohin

Grazie, i solved it.
But i have a question, (minus) which near 1/4 doesn't effect to the solution? You just left it... sorry for a crazy question.
October 26th 2008, 07:28 PM
Chris L T521

Quote:

Originally Posted by JrShohin

Grazie, i solved it.
But i have a question, (minus) which near 1/4 doesn't effect to the solution? You just left it... sorry for a crazy question.

Note that $-\frac{1}{4}\log\left|\frac{1}{4y+1}\right|=\frac{1 }{4}\log|4y+1|$

So I just used $\frac{1}{4}\log|4y+1|$ , rather than $-\frac{1}{4}\log\left|\frac{1}{4y+1}\right|$

--Chris