User:Egm6321.f10.team6.Kim.MK/hw4

= Problem 2 Verify exactness of Bessel Equation =

Given

 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} x^{2} y'' + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.1)
 * 
 * }

(a) (1) & (2) lecture note p. 15_3

 * 2nd condition of exactness

(b) (2) lecture note p. 22_4
where $$\begin{align} f_i = \frac{\partial F}{\partial y^{(i)}} ,i= 0,1,2 \end{align}$$

2) If (2.1) is not exact, see whether it can be made exact using Integrating Factor Method with $$\begin{align} h(x,y)=x^{m} y^{n}\end{align}$$

 * See lecture note p. 19_1, p.22_4

(a)
From (2.1), (2.2)

(2.6) substitutes for (2.2)

if $$\begin{align}{\nu}^2 = x^{2} +1 \end{align}$$, (2.1) is exact.

From (2.3)

(2.8) substitutes for (2.3)

This satisfies the 2nd condition of exactness.

(b)

 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} F(x,y,y',y) = x^{2} y + x y' + \left(x^{2} - {\nu}^{2} \right)y = 0 \end{align}$$

(2.10)
 * 
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} &f_0 = \frac{\partial F }{\partial y^{(0)}} = x^2 - {\nu}^2\\ &f_1 = \frac{\partial F }{\partial y^{(1)}} = x\\ &f_2 = \frac{\partial F }{\partial y^{(2)}} = x^2\\ \end{align}$$

(2.11)
 * 
 * }


 * {| style="width:100%" border="0"


 * style="width:95%" |
 * style="width:95%" |

$$\begin{align} \frac{\mathrm{d} f_1 }{\mathrm{d} x} =1 {~}{~}{~}{~}{~} \frac{\mathrm{d^2} f_2 }{\mathrm{d} x^2} =2 \end{align}$$

(2.12)
 * 
 * }

Therefore,

As same as (a), if $$\begin{align}{\nu}^2 = x^{2} +1 \end{align}$$, (2.1) is exact.

2)
Verify exactness of (2.1) using 2nd condition of exactness (2.2), (2.3)

(2.18) substitutes for (2.3)

To satisfy (2.19), n=0.

(2.16), (2.17) substitute for (2.2)

Solve the quadratic equation for m,

Thus, when $$\begin{align} m= \frac{-2 {\pm} \sqrt{{\nu}^{2}-x^{2}}}{2}, n=0\end{align}$$ (2.14) is exact.