Thread: 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.


y(0) = 0

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.


y(0) = 0

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

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.

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

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

--Chris