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Okay i've been at it for far too long now. It comes from a bigger question from working with a PDE. I did seperation of variables and now I am stuck near the end of the problem. Here is the ODE in question $$ \Phi''(y)= \lambda^2\cdot\Phi(y)$$ with the following initial condition $$ \Phi'(H)=0$$ where $H$ is a positive number. Also I know $\displaystyle \lambda = \frac{n\pi}{L} >0$. There is another condition but I dont think it can help $$ \Phi(0) = \begin{cases} 0 & x > L/2 \\ 1 & x < L/2 \end{cases}$$ sorry, i dont know how to do cases in latex and yes, that is an $x$ in the initial condition. Like i said this is a bigger problem that has both x and y.

The solution should be $$\Phi(y) = B \cdot \cosh{(\lambda(H-y))}$$

I've tried going through the following general solutions $$\Phi(y) = Ae^{\lambda y} + B e^{-\lambda y}$$ and $$\Phi(y) = A \sinh{\lambda y} + B \cosh{\lambda y}$$ but no luck that way :(

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It is probably easiest just to remember as a general fact that $$f(x)=\begin{cases}A\cosh(\lambda(x-H)) & A, H\in\mathbb R\\ A\sinh(\lambda(x-H)) & A, H\in\mathbb R\\ e^{\lambda(x-H)} & H\in\mathbb R\\ e^{-\lambda(x-H)} & H\in\mathbb R\end{cases}$$ are a complete set of solutions to $f''=\lambda^2f$ in general. To remember that they are solutions, just note that since the ODE does not mention the independent variable we can slide any known solution to the left and right. Proving them complete is slightly more involved -- one strategy would be to show that together they cover all possible pairs of $f(0), f'(0)$, by case analysis on their logarithmic derivatives.

Then, since you're already given a $H$ such that $f'(H)=0$ you can immediately see that it must be of the form $A\cosh(x-H)$ for some $A$, since the other forms have no stationary points. Finding $A$ such that your condition for $\Phi(0)$ holds is then a simple matter of scaling.

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    What does it mean by " since the ODE does not mention the independent variable we can slide any known solution to the left and right." Also, what does it mean by saying the other forms have no stationary points2011-11-26
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    The ODE is $\frac{d^2y}{dx^2}=\lambda^2 y$, and it doesn't mention $x$, only _differences_ in $x$. So if you add or subtract a constant from all $x$s (thereby moving its graph to the left or right), it cannot make a solution into a nonsolution or vice versa. -- A stationary point of a function $f$ is an $x$ such that $f'(x)=0$ There are clearly no such points for $\sinh$ or the pure exponential forms.2011-11-26
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    Thankyou. To be honest I still dont understand the "differences" part but I'll do some more research on my own now that I know what it is. Also would $x=0$ be considered a stationary point or are zeros not considered a stationary point.2011-11-26
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    The differences point is simply that $x\mapsto f(x)$ is a solution if and only of $x\mapsto f(x+c)$ is a solution. $x=0$ is a [stationary point](http://en.wikipedia.org/wiki/Stationary_point) if and only if $f'(0)=0$. It's not a deep concept -- I could just as well have said "... since $f'(x)$ is always nonzero for the other forms".2011-11-26
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    Also what does it mean by having $A, H \in \mathbb{R}$2011-11-27
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    Also in my usual cases when I solve by seperation of variables I usually get a nice answer such as $f(y) = A cosh(\lambda y)$ Is there scaling there too (just the scaling is 0?). Also is there a wiki article I can read to find out more about the solutions?2011-11-27
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    The $A,H\in\mathbb R$ are just an attempt to indicate that that each choice of $A$ and $H$ will give rise to a different solution. -- In $f(y)=A\cosh(\lambda y)$, the $A$ is a scaling factor. (Are you confusing scaling with translation?)2011-11-27
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The forms $Ae^{\lambda y} + Be^{-\lambda y}$ and $D \cosh(\lambda(C - y))$ are really different ways of writing the same thing. For $$Ae^{\lambda y} + Be^{-\lambda y} = \sqrt{AB} ( \sqrt{A \over B} e^{\lambda y} + \sqrt{B \over A} e^{-\lambda y})$$ $$ = \sqrt{AB} ({ e^{\lambda y + {1 \over 2}\ln({A \over B})} + e^{-\lambda y - {1 \over 2}\ln({A \over B})}})$$ Letting $C = -{1 \over 2\lambda}\ln{A \over B}$, this becomes $$\sqrt{AB} ( e^{\lambda y - \lambda C}+ e^{-\lambda y + \lambda C})$$ $$=2\sqrt{AB} \cosh(\lambda(y - C)) = 2\sqrt{AB} \cosh(\lambda(C - y)) $$ Thus letting $D = 2\sqrt{AB}$ you get the other form. Plug in your initial condition to show $C = H$.

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    thankyou. is there another way to do this?2011-11-26
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    you could just keep the solution in the form $Ae^{\lambda y} + Be^{-\lambda y}$. Your initial condition will give $A$ in terms of $B$ so your solutions will be $B \times$ function, which will also be correct although written a different way.2011-11-26
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    Do you have a A in terms of B so I can compare my answer? I got $A = B e^{H\lambda}$2011-11-26
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    I think it's $A = Be^{-2H\lambda}$.2011-11-27