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I have asked this question previously but I would like to address it in a new way. I am considering the equation $$ \epsilon\frac{d^2y}{dx^2}+\frac{dy}{dx}+y=0,$$ with boundary conditions $y(0)=0,y(1)=1$.

I am asked to seek a solution in the form $$y(x)=Ce^{\alpha x}\sinh(\beta x),$$ where $\alpha, \beta$ and $C$ are constants to be determined, to find the exact solution of my initial problem.

So my professor has advised to sub the given solution in to the main equation then equate coefficients of $\sinh(\beta x)$ and $\cosh(\beta x)$. I have been trying to do this but I can only ever find that $\alpha = -1$ and $\beta = 0$ which of course makes the solution redundant so there must be a problem.

Following my professors approach, note the derivatives are:

$y'=Ce^{\alpha x}(\alpha\sinh{\beta x}+\beta\cosh(\beta x))$ $y''=Ce^{\alpha x}(\alpha \beta \cosh(\beta x) + \beta^2\sinh(\beta x))+\alpha Ce^{\alpha x}(\alpha\sinh(\beta x)+\beta\cosh(\beta x))$

$\underline{\sinh(\beta x)}$

$Ce^{\alpha x}(\epsilon\beta^2+\epsilon\alpha^2+\alpha +1)$

$\underline{\cosh(\beta x)}$

$Ce^{\alpha x}(2\epsilon\alpha\beta +\beta)$

We are given initially that $0<\epsilon<<1$ so I believe we can ignore the $\epsilon$ terms which leaves us with the relation $\beta = \alpha + 1$.

Unfortunately I have no idea how to take this any further towards actually finding values for $\alpha$, $\beta$ and $C$.

My lecturer has explained that it is not that difficult to do so I believe I am not seeing something obvious here.

I also would like to ask, why did he suggest to equate the coefficients? Is it the case that for an equation of the form:

$a\sinh(bx)+c\cosh(bx)=0 \implies a=c$?

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Following on from what you have already done, you can solve $$\epsilon\beta^2+\epsilon\alpha^2+\alpha +1=0$$ and $$2\epsilon\alpha\beta +\beta=0$$ to get $$\alpha=-\frac{1}{2\epsilon}$$ and $$\beta=\pm\frac{\sqrt{1-4\epsilon}}{2\epsilon}$$

Since $\epsilon<<1$, you can make the approximation $$(1-4\epsilon)^{\frac 12}\simeq1-2\epsilon+O(\epsilon)^2$$$$\implies\beta\simeq\pm\left(\frac{1}{2\epsilon}-1\right)$$

As far as the initial conditions are concerned, the requirement that $y(0)=0$ is already satisfied. You can use the other condition to get an expression for $C$.

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  • $\begingroup$ Thank you! So based on this and considering the case where $\beta >0$ for example am I correct in saying $ C = \frac{2}{e^{-1}-e^{1-\frac{1}{\epsilon}}}$? (I had to re-edit this comment mulptiple times, apologies) $\endgroup$ – Evan Mar 4 '18 at 16:15
  • $\begingroup$ @N.K. Yes that's what I get for $C$ $\endgroup$ – David Quinn Mar 4 '18 at 16:18
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The general solution is given in the form of a linear combination of two basic, exponential solutions. So we search for solutions $y=\mathrm e^{rx}$ for some $r$. This leads ot solving $$\varepsilon r^2\mathrm e^{rx}+ r\mathrm e^{rx}+\mathrm e^{rx}=(\varepsilon r^2+r+1)\,\mathrm e^{rx}=0.$$ Thus $r$ must satisfy the characteristic equation $$\varepsilon r^2+r+1=0.$$ Its discriminant is $\;1-4\varepsilon$, which is positive if $\varepsilon <\frac14$. In this case the solutions in the characteristic equation are $$r_1,r_2=\frac{-1\pm\sqrt{1-4\varepsilon}}{2\varepsilon}$$ and the basic solutions are $$y_1(x)=\mathrm e^{\tfrac{-1+\sqrt{1-4\varepsilon}}{2\varepsilon}}\!=\mathrm e^{-\tfrac{1}{2\varepsilon}}\mathrm e^{\tfrac{\sqrt{1-4\varepsilon}}{2\varepsilon}},\qquad y_2(x)=e^{-\tfrac{1}{2\varepsilon}}\mathrm e^{-\tfrac{\sqrt{1-4\varepsilon}}{2\varepsilon}}.$$ From these basic solutions, can you deduce two basic solutions in the prescribed form?

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Hint. Note that $$y(x)=Ce^{\alpha x}\sinh(\beta x)=\frac{C}{2}\left(e^{(\alpha+\beta) x}-e^{(\alpha-\beta) x}\right)$$ Now $y(x)$ is a solution of the given homogeneous linear differential equation with constant coefficients if and only if $\alpha+\beta$ and $\alpha-\beta$ solve the characteristic equation $\epsilon z^2+z+1=0$. Hence $$2\alpha=-\frac{1}{\epsilon}\; \text{(sum of roots) and }\alpha^2-\beta^2=1/\epsilon\; \text{(product of roots)}$$ which implies, for $\epsilon\leq 1/4$ and $\epsilon\not=0$ (otherwise we have no $\beta$), $$\alpha=-\frac{1}{2\epsilon}\quad \text{and}\quad \beta=\frac{\sqrt{1-4\epsilon}}{2\epsilon}\quad \text{(by symmetry it suffices to consider just plus sign)} $$ Now determine $C$ such that $y(1)=1$ ($y(0)=0$ is already satisfied). Pay attention to the case $\epsilon=1/4$.

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  • $\begingroup$ Thank you very much, I am a little confused by the last line however. I find that $ C = \frac{2}{e^{-1}-e^{1-\frac{1}{\epsilon}}}$, when you say to consider the case $ \epsilon = \frac{1}{4}$ this implies $ C = \frac{2e^3}{e^2-1}$ would this then be the correct expression for $C$? $\endgroup$ – Evan Mar 4 '18 at 16:17
  • $\begingroup$ If $\epsilon=1/4$ then $\beta=0$ and $y(x)$ is identically zero $\endgroup$ – Robert Z Mar 4 '18 at 16:20

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