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Find the extremals of J(y)=\int _{0}^{1}(yy'+y''^{2})dx,y(0)=0,y'(0)=1,y(1)=2,y'(1)=4\,

Use the Euler-Lagrange equation for the integrand L\,

L_{y}-{\frac  {d}{dx}}L_{{y^{\prime }}}+{\frac  {d^{2}}{dx^{2}}}L_{{y^{{\prime \prime }}}}=0


y^{\prime }-{\frac  {d}{dx}}(y)+{\frac  {d^{2}}{dx^{2}}}(2y^{{\prime \prime }}) =0\,
y^{\prime }-y^{\prime }+2y^{{\prime \prime \prime \prime }}\, =0\,
y^{{\prime \prime \prime \prime }}\, =0\,
y^{{\prime \prime \prime }}\, =c_{1}\,
y^{{\prime \prime }}\, =c_{1}x+c_{2}\,
y^{{\prime }}\, ={\frac  {1}{2}}c_{1}x^{2}+c_{2}x+c_{3}\,
y\, ={\frac  {1}{6}}c_{1}x^{3}+{\frac  {1}{2}}c_{2}x^{2}+c_{3}x+c_{4}\,

Solving for the constants

y(0)=0\, \Longrightarrow c_{4}=0
y^{\prime }(0)=1\, \Longrightarrow c_{3}=1
y(1)=2\, \Longrightarrow {\frac  {c_{1}}{6}}+{\frac  {c_{2}}{2}}+1+0=2
y^{\prime }(1)=4\, \Longrightarrow {\frac  {c_{1}}{2}}+c_{2}+1=4

Thus, solving the system of equations

{\begin{cases}{\frac  {c_{1}}{6}}+{\frac  {c_{2}}{2}}=1\\{\frac  {c_{1}}{2}}+c_{2}=3\end{cases}}
\Longrightarrow c_{2}=0,c_{1}=6

Therefore, y=x^{3}+x\, is an extremal.

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