User:Egm6321.f10.team5.steinberg/hw4

Given
Bessel Differential Equation

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

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

where

$$ g(x,y,y') = xy' + (x^2 - \nu^2)y \! $$

$$ f(x,y,y') = x^2 \! $$

The second condition of exactness from p. 15-3 has two relations given here.

where

$$ p := y' \! $$

Computing the partial derivatives for (3.2a) and (3.2b),

$$ f_{xx} = 2 \! $$

$$ f_{xy} = f_{yy} = f_{xp} = f_{yp} = f_{y} = 0 \! $$

$$ g_{xp} = 1 \! $$

$$ g_{yp} = g_{pp} = 0 \! $$

$$ g_{y} = x^2 - \nu^2 \! $$

Now we substitute the derivatives into (3.2a) and (3.2b) to get

$$ (3.2a) \Rightarrow -1 = x^2 - \nu^2 \! $$

$$ (3.2b) \Rightarrow 0 = 0 \! $$

Clearly, the first relation (3.2a) 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

where $$ f_i:=\frac{\partial F}{\partial y^{(i)}}, ~i=0,1,2,...,n \! $$

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 (3.3) yields

$$ x^2 - \nu^2 - 1 + 2 = 0 \! $$

which is the same result obtained from (3.2a) 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 (3.2a) and (3.2b).

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 (3.2a) and (3.2b) yields

$$ (3.2a) \Rightarrow (m+1)(m+2)x^my^n + 2pn(m+2)x^{m+1}y^{n-1} + p^2n(n-1)x^{m+2}y^{n-2} = (m+1)x^my^n + pnx^{m+1}y^{n-1} - npx^{m+1}y^{n-1} -(n+1)x^{m+2}y^n + (n+1)x^my^n\nu^2 \! $$

$$ (3.2b) \Rightarrow 0 + 0 + 2nx^{m+2}y^{n-1} = 0 \! $$

The second relation $$ (3.2b) \Rightarrow n = 0 \! $$.

Therefore, the first relation (3.2a) simplifies to

This relation (3.4) cannot be satisfied with any value of $$ m \! $$ to make the Bessel differential equation exact.