# How can solve this differential equation (second equation)?

How can I solve this differential equation? $$\frac{1}{h}\left(\frac{a}{\delta} - \frac{b}{\delta} - \frac{c}{E\delta^4}\right) = \frac{\mathrm{d}}{\mathrm{d}x}\left[A\,\frac{\mathrm{d}\delta}{\mathrm{d}x} \left(\frac{1}{\delta} + \frac{3B}{\delta^2}\right)\right],$$ where $a,b,c,E,h,A,B$ are constants.

• no boday can help me :( – shatrah Feb 10 '14 at 8:01
• no boday can help me :( – – shatrah Feb 11 '14 at 11:52
• Where are you getting all these crazy DEs from? – IAmNoOne Feb 22 '14 at 5:36
• Where are you getting all these equations???!!! – recursive recursion Feb 22 '14 at 20:29
• Its appear on my work :( – shatrah Feb 23 '14 at 4:55

Let $y=\dfrac{d\delta}{dx}$ ,

Then $\dfrac{1}{h}\left(\dfrac{a}{\delta}-\dfrac{b}{\delta}-\dfrac{c}{E\delta^4}\right)=\dfrac{d}{d\delta}\left(Ay\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)\right)\dfrac{d\delta}{dx}$

$\dfrac{1}{h}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)=A\dfrac{d}{d\delta}\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)y$

$\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y~d\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)=\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}-\dfrac{c}{E\delta^4}\right)\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)d\delta$

$\int\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y~d\left(\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)y\right)=\int\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta^2}+\dfrac{3(a-b)B}{\delta^3}-\dfrac{c}{E\delta^5}-\dfrac{3Bc}{E\delta^6}\right)d\delta$

$\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)^2\dfrac{y^2}{2}=-\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}+\dfrac{3(a-b)B}{2\delta^2}-\dfrac{c}{4E\delta^4}-\dfrac{3Bc}{5E\delta^5}\right)+c_1$

• I still face problem to solve Eq: $$\frac{d\delta }{dx}=\sqrt{\frac{2}{\left(\dfrac{1}{\delta}+\dfrac{3B}{\delta^2}\right)^2}\left(-\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta}+\dfrac{3(a-b)B}{2\delta^2}-\dfrac{c}{4E\delta^4}-\dfrac{3Bc}{5E\delta^5}\right)+\dfrac{1}{Ah}\left(\dfrac{a-b}{\delta_o}+\dfrac{3(a-b)B}{2\delta_o^2}-\dfrac{c}{4E\delta_o^4}-\dfrac{3Bc}{5E\delta_o^5}\right) \right)}$$ where $\delta_o$ is constant – shatrah Feb 23 '14 at 6:53
• ....please could you see my comment – shatrah Feb 23 '14 at 7:03
• I still have $\dfrac{ddelta}{dx}$ as function and cant solve ? – shatrah Feb 24 '14 at 2:55