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

Solution

Verified
Step 1
1 of 2

Given:

\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 $2x$ is the derivative of $1+x^2$, which implies that we can rewrite the differential equation as:

\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 $L[y]=-(p(x)y')'+q(x)y$ and the boundary conditions are separated (that is, each condition involves only one boundary point).

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

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

## Recommended textbook solutions #### Fundamentals of Differential Equations

9th EditionArthur David Snider, Edward B. Saff, R. Kent Nagle
2,119 solutions #### Elementary Differential Equations and Boundary Value Problems

10th EditionRichard C. Diprima, William E. Boyce
3,088 solutions #### Differential Equations with Boundary-Value Problems

7th EditionDennis G. Zill, Michael R. Cullen
2,408 solutions #### A First Course in Differential Equations with Modeling Applications

10th EditionDennis G. Zill
1,989 solutions