Solve this equation in Sturm-Liouville.(1−x2)y″−xy′+λy=0My solution:μ(1−x2)y″−μxy′+μλy=(py′)′+(λr−q)yμ(1−x2)y″−μxy′+μλy=p′y′+py″+(λr−q)yp′=μx,p=μ(1−x2)⇒p′p=x1−x2⇒ln⁡(p)=−12ln|1−x2|⇒p=1(|1−x2|)Then μ=1(1−x2)31(1−x2)2y″−1(1−x2)3xy′+1(1−x2)3λy=p′y′+py″+(λr−q)y

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diocedss33 · Accepted Answer

Explanation:First, denote P(x)=1−x2;,;;Q(x)=−x⇒e∫QPdx=e12∫−2x1−x2dx=elog|1−x2|12=1−x2so your integration factor (this could in fact better be called The S-L factor...) is1Pe∫QPdx=11−x21−x2=(1−x2)−12Multiply now the whole diff. equation by it and get:1−x2,y″−x1−x2y′+λ11−x2y=0⇔((1−x2)12y′)′+λ1−x2,y=0and there you have your diff. eq. in Sturm-Liouville normal form.

Solve this equation in Sturm-Liouville. \(\displaystyle{\left({1}-{x}^{{2}}\right)}{y}{''}-{x}{y}'+\lambda{y}={0}\) My solution: \(\displaystyle\mu{\left({1}-{x}^{{2}}\right)}{y}{''}-\mu{x}{y}'+\mu\lambda{y}={\left({p}{y}'\right)}'+{\left(\lambda{r}-{q}\right)}{y}\)

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