<html><head><meta name="color-scheme" content="light dark"></head><body><pre style="word-wrap: break-word; white-space: pre-wrap;">\documentclass[a4paper,12pt]{article}
\newcommand{\ds}{\displaystyle}
\newcommand{\un}{\underline}
\newcommand{\undb}{\underbrace}
\newcommand{\pl}{\partial}
\parindent=0pt
\begin{document}
In what follows you may assume that the following notation applies

$$y=y(x),\ y'=\ds\frac{dy}{dx}.$$

You may also assume that, unless otherwise stated, $y$ is a
sufficiently continuously differentiable function.

{\bf Question}

Find $y(x)$ such that

$$I=\ds\int_0^a \,dx (y^2+2{y'}^2+{y''}^2)$$

is stationary and $y(0)=1+\ds\frac{5a}{4},\
y(a)=\ds\frac{5}{3}+\ds\frac{3a}{4},\ y'(0)=0,\ y'(a)=0$ where
$a=\log 3$.

\medskip

{\bf Answer}

The relevant E-L equation is (generalistion 1)

$$\ds\frac{d^2}{dx^2}(2y'')-\ds\frac{d}{dx}(4y')+2y=0$$

which gives

$$y^{iv}-2y''+y=0$$

From lectures it is convenient to write the solution as

$$y=A\cosh x+B\sinh x+Cx\cosh x+Dx \sinh x$$

$y'(0)=0 \Rightarrow 0=B+C$

$y(0)=1+\ds\frac{5\log 3}{4} \Rightarrow A=1+\ds\frac{5}{4}\log 3$

Since at $x=\log 3,\ \cosh x=\ds\frac{5}{3},\ \sinh
x=\ds\frac{4}{3}$ we have that the 2 remaining conditions
$\Rightarrow B=0$ (and so $C=0$), and $D=-1$. Hence $y=A \cosh
x-x\sinh x$, where $A=1+\ds\frac{5}{4}\log 3$.

\end{document}
</pre></body></html>