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Minimize J(y)=\int _{0}^{\infty }\left(y^{2}+y'^{2}+(y''+y')^{2}\right)dx,y(0)=1,y'(0)=2,y(\infty )=0,y'(\infty )=0\,

Let F(x,y,y',y'')\, denote the integrand in the above. The Euler equation (see CoV4) for this problem is

{\frac  {d^{2}}{dx^{2}}}\left({\frac  {\partial F}{\partial y''}}\right)-{\frac  {d}{dx}}\left({\frac  {\partial F}{\partial y'}}\right)+{\frac  {\partial F}{\partial y}}=0

\implies {\frac  {d^{2}}{dx^{2}}}2(y''+y')-{\frac  {d}{dx}}[2y'+2(y''+y')]+2y=0

\implies y^{{(iv)}}-2y''+y=0

The characteristic equation for this differential equation is m^{4}-2m^{2}+1=0 which factors to (m-1)^{2}(m+1)^{2}=0; so the roots are m=1,1,-1,-1. Thus, the general solution of the differential equation is y=(c_{1}+c_{2}x)e^{x}+(c_{3}+c_{4}x)e^{{-x}}\,

The conditions at infinity require that c_{1}=c_{2}=0\,. Thus, y=(c_{3}+c_{4}x)e^{{-x}},\quad y'=\left((c_{4}-c_{3})-c_{4}x\right)e^{{-x}}\,

Applying the remaining conditions,

c_{3}=1,\quad c_{4}-c_{3}=2\, so c_{3}=1,c_{4}=3\, and so the desired solution is


Satisfying the Euler equation is a necessary condition for the given solution to minimize J(y)\,; proving that it actually does minimize J(y)\, is a lot more difficult (and will not be dealt with here).

Calculus of Variations

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