\( \rightarrow \) Find Whe sourin of Revcci founa \[ x^{2} \frac{d^{2} y}{d x}+x \frac{d y}{d x}+\left(x^{2}-p^{2}\right) y=0 \] bole (17) As (giren) \[ \begin{array}{l} \Rightarrow x^{2} y^{\prime \prime}+x y^{\prime}+\left(x^{2}-y^{2}\right) y=0-\infty \\ \because x\left(x y^{\prime}\right)^{\prime}=x\left[y^{\prime}+x y^{\prime \prime}\right]=x y^{\prime}+x^{\prime} y^{\prime \prime} \end{array} \] Hence: \[ \text { (1) } \Rightarrow x\left(x y^{\prime}\right)^{\prime}+\left(x^{2}-p^{2}\right) y=0 \] Let. \[ \begin{array}{l} y=\sum_{n=0}^{\infty} a_{n} x^{n+s} \\ \Rightarrow y^{\prime}=\sum_{n \rightarrow 0}(n+s) a_{n} x^{n+s-1} \\ \Rightarrow\left(x y^{\prime}\right)=\sum_{n=0}^{\infty}(n+s) a_{n} x^{n+s} \text {. } \\ \Rightarrow\left(x y^{\prime}\right)^{\prime}=\sum_{n=0}(n+s)^{2} a_{n} x^{n+s-1} \\ =x\left(n y^{\prime}\right)^{\prime}=\sum_{n=0}^{n}(n+s)^{2} a_{n} x^{n+s} \text {. } \\ \text { 4. } x^{2} y=\sum_{n=0}^{m} a_{n} x^{n+s+2} \\ \text { (4) } \rightarrow\left(-p^{2} y=-\frac{1}{n=0} a_{n} x^{n+5} \cdot p^{2}\right. \\ \end{array} \] Table: Get:
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Step 1: Start with the given Bessel's differential equation: \[ x^2 \frac{d^2 y}{dx^2} + x \frac{dy}{dx} + (x^2 - p^2)y = 0 \] Show moreβ¦
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$\left(\frac{x}{a}\right)^{n}+\left(\frac{y}{b}\right)^{n}=2$ $\frac{n}{a}\left(\frac{x}{a}\right)^{n-1}+\frac{n}{b}\left(\frac{y}{b}\right)^{n-t} \frac{d y}{d x}=0$ $\frac{d y}{d x}=-\frac{b^{n} x^{n-1}}{a^{n} y^{n-1}}=-\frac{b}{a}$
$$x^{2} y^{n}+x y^{\prime}+x^{2} y=0$$
Series Solutions of Differential Equations
Method of Frobenius
$\begin{aligned} &x^{2}+y^{2}=a^{2} \\ &2 x+2 y y^{\prime}=0 \Rightarrow \frac{x}{y}=-y^{\prime} \\ &1+\left(y^{\prime}\right)^{2}+y y^{\prime \prime}=0 \Rightarrow y=\frac{-\left(1+\left(y^{\prime}\right)^{2}\right)}{y^{\prime \prime}} \end{aligned}$ Using, (I), (II) \& (III) $\begin{aligned} &\left(\mathrm{y}^{\prime}\right)^{2}+1=\frac{\mathrm{a}^{2}\left(\mathrm{y}^{\prime \prime}\right)^{2}}{\left(1+\left(\mathrm{y}^{\prime}\right)^{2}\right)^{4}} \Rightarrow \frac{1}{\mathrm{a}^{2}}=\frac{\left(\mathrm{y}^{\prime \prime}\right)^{2}}{\left(1+\left(\mathrm{y}^{\prime}\right)^{2}\right)^{2}\left(1+\left(\mathrm{y}^{\prime}\right)^{2}\right)} \\ &\Rightarrow \mathrm{K}=\frac{\left|\mathrm{y}^{\prime \prime}\right|}{\sqrt{\left(1+\left(\mathrm{y}^{\prime}\right)^{2}\right)^{3}}} \end{aligned}$
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