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

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- Oct 26th 2008, 06:40 AMJrShohinAnother 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 - Oct 26th 2008, 06:48 AMChris L T521
I would leave it as $\displaystyle \frac{1}{4}\log\left|4y+1\right|=\frac{x^2}{2}+C$

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

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

Now solve for y. Then apply the initial condition.

Can you take it from here?

--Chris - Oct 26th 2008, 07:13 AMJrShohin
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. - Oct 26th 2008, 07:28 PMChris L T521