User:Egm6321.f2010.team5.riveros/hw4p2

= Problem 2 - Solve for Bessel equation = From Meeting 24, p. 24-1

Given
Bessel Differential Equation
 * $$ \displaystyle F(x,y,y',y) = x^2y + xy' + (x^2-\nu^2)y,  ~\nu \in\mathbb{R}  \! $$.

Find

 * 1) Verify exactness of the given equation using 2 methods:
 * 2) Relations (1) and (2) p. 15-3 (Meeting 15)
 * 3) Equation (3) p. 22-4 (Meeting 22)
 * 4) If not exact, determine whether it can be made exact using integrating factor method (IFM) with $$ h(x,y) = x^my^n \! $$

Part 1
The first condition of exactness is satisfied since the Bessel differential equation is in the form
 * {| style="width:100%" border="0"

where,
 * style="width:95%" | $$\displaystyle F(x,y,y',y) = g(x,y,y') + f(x,y,y')y$$
 * (2.1)
 * }
 * }
 * $$\displaystyle g(x,y,y') = xy' + (x^2 - \nu^2)y \! $$
 * $$ f(x,y,y') = x^2 \! $$

The second condition of exactness from p. 15-3 has two relations given here.
 * {| style="width:100%" border="0"


 * style="width:95%" | 1. $$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y$$
 * (2.2a)
 * }
 * {| style="width:100%" border="0"
 * {| style="width:100%" border="0"

where
 * style="width:95%" | 2. $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp}$$
 * (2.2b)
 * }
 * }
 * $$\displaystyle p := y'$$

Computing the partial derivatives for (3.2a) and (3.2b),
 * $$\displaystyle f_{xx} = 2$$
 * $$\displaystyle f_{xy} = f_{yy} = f_{xp} = f_{yp} = f_{y} = 0$$
 * $$\displaystyle g_{xp} = 1$$
 * $$\displaystyle g_{yp} = g_{pp} = 0$$
 * $$\displaystyle g_{y} = x^2 - \nu^2$$

Now we substitute the derivatives into (2.2a) and (2.2b) to get
 * (2.2a) $$\displaystyle \Rightarrow -1 = x^2 - \nu^2 \! $$
 * (2.2b) $$\displaystyle \Rightarrow 0 = 0 \! $$

Clearly, the first relation (2.2a) indicates that the Bessel differential equation is not exact for all $$x$$. Alternatively, the second condition of exactness can also be verified using (3) p.22-4 which is as follows,
 * {| style="width:100%" border="0"

where,
 * style="width:95%" | $$\displaystyle f_0 - \frac{df_1}{dx} + \frac{d^2f_2}{dx^2} = 0$$
 * (2.3)
 * }
 * }
 * $$\displaystyle f_i:=\frac{\partial F}{\partial y^{(i)}}, ~i=0,1,2,...,n$$

For the given equation,
 * $$ f_{0} = \frac{\partial F}{\partial y^{(0)}} = x^2 - \nu^2$$
 * $$ f_{1} = \frac{\partial F}{\partial y^{(1)}} = x ~\Rightarrow ~\frac{df_1}{dx} = 1$$
 * $$ f_{2} = \frac{\partial F}{\partial y^{(2)}} = x^2 ~\Rightarrow ~\frac{df_2}{dx} = 2$$

Substituting these results into (2.3) yields,
 * $$\displaystyle x^2 - \nu^2 - 1 + 2 = 0$$

which is the same result obtained from (2.2a) above. Again, the Bessel equation is not exact for all $$x$$.

Part 2
We will now employ the integrating factor method using $$h(x,y) = x^my^n$$ to see if the given equation can be made exact.
 * $$ h(x,y)\Big(x^2y'' + xy' + (x^2-\nu^2)y \Big) = 0 \! $$
 * $$ x^my^n \Big(x^2y'' + xp + (x^2-\nu^2)y \Big) = 0 \! $$

Using the form of (3.1) yields,
 * $$ g(x,y,p) = x^{m+1}y^n \! $$
 * $$ f(x,y,p) = x^{m+1}y^np + x^{m+2}y^{n+1} - x^my^{n+1}\nu^2 = 0 \! $$

To satisfy the 2nd condition of exactness, $$f(x,y,p)$$ and $$g(x,y,p)$$ must satisfy (2.2a) and (2.2b). The partial derivatives for these 2 equations are;
 * $$ f_{xx} = (m+1)(m+2)x^m y^n \! $$
 * $$ f_{xy} = n(m+2)x^{m+1} y^{n-1} \! $$
 * $$ f_{y} = nx^{m+2} y^{n-1} \! $$
 * $$ f_{yy} = n(n-1)x^{m+2} y^{n-2} \! $$
 * $$ f_{xp} = f_{yp} = 0 \! $$
 * $$ g_{xp} = (m+1)x^m y^n \! $$
 * $$ g_{y} = nx^{m+1}y^{n-1}p + (n+1)x^{m+2}y^n - (n+1)x^m y^n \nu^2 \! $$
 * $$ g_{yp} = nx^{m+1}y^{n-1} \! $$
 * $$ g_{pp} = 0 \! $$

Substituting these partial derivatives into (2.2a) and (2.2b) yields
 * (2.2a) $$ \Rightarrow (m+1)(m+2)x^my^n + 2pn(m+2)x^{m+1}y^{n-1} + p^2n(n-1)x^{m+2}y^{n-2} = (m+1)x^my^n + pnx^{m+1}y^{n-1} - npx^{m+1}y^{n-1} -(n+1)x^{m+2}y^n + (n+1)x^my^n\nu^2 \! $$
 * (2.2b) $$ \Rightarrow 0 + 0 + 2nx^{m+2}y^{n-1} = 0 \! $$

The second relation (2.2b) ⇒ $$n = 0$$. Therefore, the first relation (2.2a) simplifies to:
 * {| style="width:100%" border="0"

This relation (2.4) cannot be satisfied with any value of $$m$$ to make the Bessel differential equation exact.
 * style="width:95%" | $$\displaystyle (m+1)(m+2)x^m = (m+1)x^m -x^{m+2} + x^m\nu^2$$
 * (2.4)
 * }
 * }