Question

determine whether the given boundary value problem is self-adjoint. (1+x2)y''+2xy'+y=0,y'(0)=0,y(1)+2y'(1)=0

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Given:

L[y]=(1+x2)y+2xy+y=0y(0)=0y(1)+2y(1)=0\begin{align*} L[y]=(1+x^2)y''+2xy'+y&=0 \\ y'(0)&=0 \\ y(1)+2y'(1)&=0 \end{align*}

First, we note that 2x2x is the derivative of 1+x21+x^2, which implies that we can rewrite the differential equation as:

L[y]=(1+x2)y+2xy+y=[(1+x2)y]+y=[(1x2)y]+y\begin{align*} L[y]&=(1+x^2)y''+2xy'+y \\ &=[(1+x^2)y']'+y \\ &=-[(-1-x^2)y']'+y \end{align*}

A second-order differential equation is self-adjoint when the differential equation is of the form L[y]=(p(x)y)+q(x)yL[y]=-(p(x)y')'+q(x)y and the boundary conditions are separated (that is, each condition involves only one boundary point).

p(x)=1x2q(x)=1\begin{align*} p(x)&=-1-x^2 \\ q(x)&=1 \end{align*}

Since L[y]=(p(x)y)+q(x)yL[y]=-(p(x)y')'+q(x)y holds in this case and since the boundary conditions are separated (as one condition involves only x=0x=0 and the other condition only involves x=1x=1), the boundary value problem is self-adjoint\textbf{self-adjoint}.

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