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求解二阶微分方程\(x (x-1) y''+3 xy'+y=0\)
我的一点想法：
\(x=0\) is a regular singular point, so seek a solution as \(y (x)=\sum_{k=0}^∞a_k x^{α+k}, a_0≠0\)\[\sum_{k=0}^∞(α+k) (α+k-1) a_k x^{α+k}-\sum_{k=0}^∞(α+k) (α+k-1) a_k x^{α+k-1}+\sum_{k=0}^∞3 (α+k) a_k x^{α+k}+\sum_{k=0}^∞a_k x^{α+k}=0\]
Pull out the first term \(- α (α-1) a_0 x^{α-1}\)
\[\underbrace{- α (α-1)}_{\substack{\text{indicial equation}\\ F (α)=α (α-1)=0\\⇒α=0, 1}} a_0 x^{α-1}+\sum_{k=0}^∞\underbrace{\{ [(α+k) (α+k-1)+3 (α+k)+1] a_k-(α+k+1) (α+k) a_{k+1} \}}_{(α+k+1)^2 a_k-(α+k+1) (α+k) a_{k+1}} x^{α+k}\]With \(α=1\), we have recurrence relation
\[(α+k) a_{k+1}=(α+k+1) a_k \quad \text{for } k \geqslant 0⇒a_{k+1}=\frac{k+2}{k+1} a_k⇒a_k=(k+1) a_0\]
So \(y_1 (x)=a_0 x \sum_{k=0}^∞(k+1) x^k=a_0 \frac{x}{(1-x)^2}\)
With \(α=0\), the recurrence relation \(ka_{k+1}=(k+1) a_k\) for \(k \geqslant 0\)
但是对于 \(k=0\) 这会出错: 递归方程变为\(0a_1=a_0\)
我接下来要做什么? |
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