CoV12

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Find an extremal for J(y)=\int _{1}^{2}{\frac  {{\sqrt  {1+y'^{2}}}}{x}}dx,y(1)=0,y(2)=1\,


Since L\, is indepedent of y\,

{\frac  {\partial L}{\partial y^{\prime }}} =c_{1}\,
{\frac  {y^{\prime }}{x{\sqrt  {1+y^{{\prime 2}}}}}} =c_{1}\,
y^{\prime }\, =c_{1}x{\sqrt  {1+y^{{\prime 2}}}}
y^{{\prime 2}}\, =c_{1}^{2}x^{2}(1+y^{{\prime 2}})
y^{\prime }\, =\pm {\frac  {c_{1}x}{{\sqrt  {1-c_{1}^{2}x^{2}}}}}

So y=\pm \left({\frac  {{\sqrt  {1-c_{1}^{2}x^{2}}}}{c_{1}}}+c_{2}\right). Solving for the constants gives

{\begin{cases}0&=\pm {\frac  {{\sqrt  {1-c_{1}^{2}}}}{c_{1}}}+c_{2}\\1&=\pm {\frac  {{\sqrt  {1-4c_{1}^{2}}}}{c_{1}}}+c_{2}\end{cases}}
{\begin{cases}c_{1}&=-{\frac  {1}{{\sqrt  {5}}}}\\c_{2}&=2\end{cases}}

Therefore, an extremal is y=2-{\sqrt  {5-x^{2}}}.


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