Looking for help with solving an ODE with Laplace Transform. I know what I'm doing and I have the solution but I think a huge step was skipped and I don't understand how they got what they got.

Question

anonymous · Answer

ok

anonymous · Answer

you familiar with ?

kinggeorge · Answer

Could you post the solution you have, or at least the part where you think a step was skipped?

anonymous · Answer

Or the question itself?

anonymous · Answer

The DE: $$EI \frac{ d^{4}y }{ dx^{4} } = W_{0} \delta(x-L/2);$$ $$y(0) = 0; y'(0) = 0; y''(L) = 0; y'''(L) = 0$$ The solution: http://img694.imageshack.us/img694/848/solution13m.jpg I don't know how to use y''(L) and y'''(L)

anonymous · Answer

number 13

kinggeorge · Answer

What is $P_0$?

anonymous · Answer

Wo

anonymous · Answer

different textbook editions lol

kinggeorge · Answer

And the $\delta$ is the dirac delta?

anonymous · Answer

yes

anonymous · Answer

i just don't know how to use y''(L) and y'''(L) to get what he got

kinggeorge · Answer

I think, they just integrated 0 with respect to s. So $$\int y^{(4)} ds=y^{(3)}+C$$And since $y^{(4)}=0$ except at L/2, and $y^{(3)}(L)=0$, we have that $y^{(3)}(0)=W_0/EI$. 

Or something like that. And if you do it again, you get that $y''(0)=W_0L/2EI$.

kinggeorge · Answer

This seems a bit hand-wavy to me, but I think they've done something like that.

anonymous · Answer

Why is $$y^{(4)} = 0$$  except at L/2 ?

anonymous · Answer

oh. because of the delta dirac function

anonymous · Answer

at L/2 its inf

kinggeorge · Answer

Basically.

kinggeorge · Answer

And since you're multiplying by $W_0/EI$, when you integrate, you get $W_0/EI$ I think. I'm still not sure why you have to divide by 2 when you do it again.

anonymous · Answer

thanks for the help :) king indeed

kinggeorge · Answer

You're welcome.