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March 4th, 2019, 01:14 AM   #1
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Question Help with differential equation

Hi, I'm new to this forum. Found this forum searching for help with a Math question. I need some help with a differential equation that I can't solve. I hope you guys can help me.

The equation
$\displaystyle \frac{d^2F}{dX^2}=-\frac{2\ast\sigma_s}{E\ast\left[\sqrt{4.65152d^2-26.853\frac{M}{\sigma_sd}-0.41955d}\right]}$

With the following replacements

$\displaystyle M=\frac{P}{2}\ast x$

$\displaystyle a=4.65152d^2$

$\displaystyle b=\frac{26.853}{2}\frac{P}{\sigma_s\ast d}$

$\displaystyle c=2\frac{\sigma_s}{E}$

$\displaystyle e=0.41955d$

We can rewrite the equation as: $\displaystyle \frac{d^2F}{dx^2}=\frac{c}{e-\sqrt{a-bX}}$

Boundary conditions: $\displaystyle x=\frac{L}{2}$, $\displaystyle \frac{dF}{dX}=0$

Solving this:

$\displaystyle dF/dx=\frac{2c\ast\left(e\ast\ln{\left(\sqrt{a-bx}\ast e\right)}+\sqrt{a-bx}\right)}{b}+C_1$

$\displaystyle C_1=-\frac{2c\ast\left(e\ast\ln{\left(\sqrt{a-b\frac{L}{2}\ }\ast e\right)}+\sqrt{a-b\frac{L}{2}}\right)}{b}$

Next we say the boundary conditions are:

$\displaystyle x=x_0$

$\displaystyle F=F_0=\frac{P}{d^4\ast E}\ast(\frac{4}{\pi}L^2\ast\ x_0-\frac{16}{3\pi}\ast\ x_0^3)$

with $\displaystyle x_0=\pi/16\ast\sigma_s/P\ast d^3$

This is the point where I don't know how to solve it anymore. I know what the solution is, but I want to know how they got there.


$\displaystyle F=2\frac{c}{b}\left\{-\frac{2}{3}\frac{\left(a-bX\right)^\frac{3}{2}}{b}-\frac{e}{b}\left[ln\left(\sqrt{a-bX}-e\right)\left(a-bX-e^2\right)+\frac{3}{2}e^2-e\sqrt{a-bX}-\frac{a-bX}{2}\right]\ \right\}+C_1X+C_2$


$\displaystyle C_1=-\frac{2c}{b}\left[\sqrt{a-b\frac{L}{2}}+e*ln\left(\sqrt{a-b\frac{L}{2}}-e\right)\right]$

$\displaystyle C_2=Y_0-\frac{2c}{b}\left\{-\frac{2}{3}\frac{\left(a-bX\right)^\frac{3}{2}}{b}-\frac{e}{b}\left[\ln{\left(\sqrt{a-bX_0}-e\right)\left(a-bX_0-e^2\right)+\frac{3}{2}e^2-e\sqrt{a-bX_0}-\frac{a-bX_0}{2}}\right]\right\}-K_1X_0$


Many thanks in advance.
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Last edited by skipjack; March 4th, 2019 at 04:43 AM.
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