# Differential Equation needs checking

• Mar 19th 2011, 12:56 PM
Turlock
Differential Equation needs checking
Hi can I get a check on my working here please.

$\displaystyle \frac{dy}{dx} = \frac{2(e^x - e^{-x})}{y^2(e^x + e^{-x})^4} (y>0)$

So I'm using the separation of variables method here.

$\displaystyle y^2 \frac{dy}{dx} = 2(\frac{e^x - e^{-x}}{(e^x + e^{-x})^4})$

$\displaystyle \int y^2 dy = 2 \int (\frac{e^x - e^{-x}}{(e^x + e^{-x})^4})dx$

$\displaystyle \frac{1}{3}y^3 = -\frac{2}{3} (e^x + e^{-x})^{-3} +c$

Using the initial condition $\displaystyle y=\frac{1}{2}$ when $\displaystyle x=0$

$\displaystyle c=\frac{1}{8}$

Solving for an explicit solution I get

$\displaystyle y =\sqrt[3]{\frac{3}{8} - \frac{2}{(e + e^{-1})^3}}$
• Mar 19th 2011, 01:00 PM
e^(i*pi)
Your method seems fine. What happened to the 3 in "2/3" for your final solution though?
• Mar 19th 2011, 01:46 PM
Turlock
Quote:

Originally Posted by e^(i*pi)
Your method seems fine. What happened to the 3 in "2/3" for your final solution though?

Well it's 1/3y^3= to start with, so multiply both sides by 3.
• Mar 19th 2011, 01:47 PM
e^(i*pi)
As yes, I missed that. In that case your answer is fine