First use (20) to express the general solution of the given differential equation in terms of Bessel functions. Then use (26) and (27) to express the general solution in terms of elementary functions. x2y+4xy+(x2+2)y=0x^{2} y^{\prime \prime}+4 x y^{\prime}+\left(x^{2}+2\right) y=0


Answered 2 years ago
Answered 2 years ago
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In Section the following equations are given, which we will use in the Exercise:

y+12axy+(b2c2x2c2+a2p2c2x2)y=0,p0y=xa[c1Jp(bxc)+c2Yp(bxc)]\begin{align*} &y''+\dfrac{1-2a}{x}y'+\left(b^2c^2 x^{2c-2}+\dfrac{a^2-p^2c^2}{x^2}\right)y=0\, , \hspace*{8mm} p\geq 0 \tag{20} \\ &y=x^a\left[c_1J_p(bx^c) +c_2Y_p(bx^c)\right] \tag{21}\\ \end{align*}

We will also use the following equations

J12(x)=2πxsinxJ12(x)=2πxcosx\begin{align*} J_{\frac{1}{2}}(x)&=\sqrt{\dfrac{2}{\pi x}} \sin x \tag{26} \\ J_{-\frac{1}{2}}(x)&=\sqrt{\dfrac{2}{\pi x}} \cos x \tag{27} \\ \end{align*}

From the Exercise we have a given differential equation

x2y+4xy+(x2+2)y=0y+4xx2y+x2+2x2y=0y+4xy+(1+2x2)y=0\begin{align*} x^2y''+4xy'+\left(x^2+2\right)y&=0\\ y''+\frac{4x}{x^2}\cdot y'+\dfrac{x^2+2}{x^2}\cdot y&=0\\ y''+\frac{4}{x}\cdot y'+\left(1+\frac{2}{x^2}\right)y&=0\\ \end{align*}

We compare the given differential equation with equation (20)(20), and we can conclude that:

12a=4b2c2=12c2=0a2p2c2=2\begin{align*} 1-2a&=4\\ b^2c^2&=1\\ 2c-2&=0\\ a^2-p^2c^2&=2\\ \end{align*}

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