 Solution

Take an admissible variation $y+\epsilon \eta$ with $\eta ={\eta }^{\prime }=0$ at both regular endpoints. Its first variation is $\delta I=\int (f_y\eta +f_{y^{\prime }}{\eta }^{\prime }+f_{y^{\prime \prime }}{\eta }^{\prime \prime })dx$. Integrating the second term once and the third twice gives the boundary term $[f_{y^{\prime }}\eta +f_{y^{\prime \prime }}{\eta }^{\prime }-(d{f}_{y^{\prime \prime }}/dx)\eta {]}_a^b$, which vanishes. The fundamental lemma of the calculus of variations therefore yields the higher-order Euler-Lagrange equationHere $f_y=8y{y}^{\prime }$, $f_{y^{\prime }}=4y^2$, and $f_{y^{\prime \prime }}=2x^4{y}^{\prime \prime }$. The first two contributions cancel, leaving $(x^4{y}^{\prime \prime }{)}^{\prime \prime }=0$. For $x>0$, integrate twice to obtain $x^4{y}^{\prime \prime }=Ax+B$, and thenThe two singular-endpoint limits give $B=0$ and $A=2$. The conditions at one give $C=1$ and $D=0$. Hence the unique solution of the stated differential equation and limiting data isThere is a genuine distinction between solving this equation and stationarity of a finite improper action. Substitution gives $f=4x^2+4-4x^{-4}$, whose integral at zero diverges to $-\infty$. The displayed Euler-Lagrange equation remains valid for compactly supported variations on $(0,1)$, or on regular cut-off intervals, but the literal unregularized action is not a finite stationary functional on this solution. These singular endpoints in higher-order variational problems require a regularization or an admissible-space convention if a global variational claim is intended.