User:Egm6321.f10.team5/hw4

= Problem 1 - Solve for L2-ODE-VC = From Meeting 22, p. 22-3

Given

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 * style="width:95%" | $$\displaystyle  F = (cos \ x) y'' + (x^2 - sin \ x)y' + 2xy = 0$$
 * (1.1)
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Find

 * 1) Show (1.1) is exact
 * 2) Find $$\Phi$$
 * 3) Solve for $$y(x)$$

1. Show (1.1) is exact
1st Condition of Exactness

from Mtg 15 (2) p. 15-2
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 * style="width:95%" | $$\displaystyle F = \underbrace{(cos \ x)}_{f(x,y,p)} y'' + \underbrace{(x^2 - sin \ x)y' + 2xy}_{g(x,y,p)} = 0$$
 * (1.2)
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(1.1) satisfies the 1st Condition of Exactness
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$$\displaystyle f \ = \ cos \ x $$

$$\displaystyle f_{xx} \ = \ -cos \ x $$

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

$$\displaystyle g \ = \ (x^2 - sin \ x) p + 2xy $$

$$\displaystyle g_{xp} \ = \ 2x - cos \ x $$

$$\displaystyle g_{y} \ = \ 2x $$

$$\displaystyle g_{yp} \ = \ g_{pp} \ = \ 0 $$
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$$\displaystyle \underbrace{f_{xx} + 2pf_{xy} + p^2 f_{yy}}_{-cos \ x} \ = \ \underbrace{g_{xp} + p g_{yp} - g_{y}}_{-cos \ x} $$

$$\displaystyle \underbrace{f_{xp} + p f_{yp} + 2 f_{y}}_{0} \ = \ \underbrace{g_{pp}}_{0} $$


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(1.1) satisfies the 2nd Condition of Exactness
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Therefore, (1.1) is Exact
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2. Find $$\Phi$$
from Mtg 17 (1) p. 17-3


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$$\displaystyle \phi \ = \ h(x,y) + \underbrace{\int f(x,y,p) \ dp}_{p \ cos \ x} $$
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$$\displaystyle \phi_{x} \ = \ h_{x} - p \ sin \ x $$

$$\displaystyle \phi_{y} \ = \ h_{y} $$

$$\displaystyle g \ = \ (h_{x} - p \ sin \ x) + h_{y} \ p \ = \ (h_{y} - sin \ x)p + h_{x} $$


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Compare to (1-2), we can choose easily


 * $$\displaystyle h_{y} = x^2$$


 * $$\displaystyle h_{x} = 2xy$$


 * $$\displaystyle \therefore h = x^2 y$$


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 * style="width:95%" | $$\phi \ = \ x^2 y + p \ cos \ x  \ = \ \underbrace{k}_{constant}$$
 * (1.3)
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3. Solve for $$y(x)$$
(1.3) -->


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 * style="width:95%" | $$\displaystyle y' + \frac{x^2 y}{cos \ x} \ = \ \frac{k}{cos \ x}$$
 * (1.4)
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$$h \ = \ exp \int_{}^{x} \frac{s^2}{cos \ s} \ ds$$
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$$y(x) \ = \ \frac{1}{exp \int_{}^{x} \frac{s^2}{cos \ s} \ ds} \ \int_{}^{x} (exp  \int_{}^{s} \frac{t^2}{cos \ t} \ dt) \ \frac{1}{cos \ s} \ ds$$
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Author and proof-reader
[Author]
 * Oh, Sang Min

[proof-reader]
 * Mike Faraone

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

Given
We are given the Bessel differential equation,
 * $$ \displaystyle F(x,y,y',y) = x^2y + xy' + (x^2-\nu^2)y,  ~\nu \in\mathbb{R}  \! $$.

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

Part 1
The first condition of exactness is satisfied since the Bessel differential equation is in the form
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where,
 * style="width:95%" | $$\displaystyle F(x,y,y',y) = g(x,y,y') + f(x,y,y')y$$
 * (2.1)
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 * $$\displaystyle g(x,y,y') = xy' + (x^2 - \nu^2)y \! $$
 * $$ f(x,y,y') = x^2 \! $$

The second condition of exactness from p. 15-3 has two relations given here.
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 * style="width:95%" | 1. $$\displaystyle f_{xx} + 2pf_{xy} + p^2f_{yy} = g_{xp} + pg_{yp} - g_y$$
 * (2.2a)
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where
 * style="width:95%" | 2. $$\displaystyle f_{xp} + pf_{yp} + 2f_y = g_{pp}$$
 * (2.2b)
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 * }
 * $$\displaystyle p := y'$$

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

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

For the given equation,
 * $$ f_{0} = \frac{\partial F}{\partial y^{(0)}} = x^2 - \nu^2$$
 * $$ f_{1} = \frac{\partial F}{\partial y^{(1)}} = x ~\Rightarrow ~\frac{df_1}{dx} = 1$$
 * $$ f_{2} = \frac{\partial F}{\partial y^{(2)}} = x^2 ~\Rightarrow ~\frac{d^2f_2}{dx^2} = 2$$

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

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) $$\displaystyle \Rightarrow (m+1)(m+2)x^my^n + 2pn(m+2)x^{m+1}y^{n-1} + p^2n(n-1)x^{m+2}y^{n-2}$$
 * $$=\displaystyle (m+1)x^my^n + pnx^{m+1}y^{n-1} - npx^{m+1}y^{n-1} -(n+1)x^{m+2}y^n + (n+1)x^my^n\nu^2$$
 * (2.2b) $$\displaystyle \Rightarrow 0 + 0 + 2nx^{m+2}y^{n-1} = 0 \! $$

The second relation (2.2b) ⇒ $$n = 0$$. Therefore, the first relation (2.2a) simplifies to:
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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)
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Author and proof-reader
[Author]


 * Michael Steinberg

[proof-reader]


 * Raul Riveros


 * Rob Carroll


 * Mike Faraone

= Problem 3 - Find $$X(x)$$= From Meeting 25, p. 25-2

Given
The following identities are given,
 * $$\displaystyle \exp (i \theta) = \cos (\theta) + i \sin (\theta)$$

and
 * $$\displaystyle \exp (\theta) = \cosh (\theta) + \sinh (\theta)$$.

Find
Find $$X(x)$$ in terms of $$\sin$$, $$\cos$$, $$\sinh$$, and $$\cosh$$.

Solve
Assume that $$X(x) = \exp(rx)$$. From the equation of $$K^4 X^{(4)} = X$$, it was already derived in the lecture notes that
 * $$r_{1,2} = \pm \frac 1 K$$

and
 * $$r_{3,4} = \pm \frac i K$$.

Since there are four solutions for $$r$$, we make a linear combination to express $$X(x)$$.
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 * style="width:95%" | $$\displaystyle X(x) = c_1 \exp (r_1 x) + c_2 \exp (r_2 x) + c_3 \exp (r_3 x) + c_4 \exp (r_4 x)$$
 * (3.1)
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Substitute the values of $$r$$s into the equation and we get
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 * style="width:95%" | $$\displaystyle X(x) = c_1 \exp (\frac x K) + c_2 \exp (- \frac x K) + c_3 \exp (i \frac x K) + c_4 \exp (- i \frac x K)$$
 * (3.2)
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Now use the expression of exponential in terms of trigonometric functions.
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 * style="width:95%" | $$\displaystyle X(x) = c_1 \left[\cosh(\frac x K) + \sinh (\frac x K)\right] + c_2 \left[\cosh(- \frac x K) + \sinh (- \frac x K)\right]+ c_3 \left[ \cos (\frac x K) + i \sin (\frac x K) \right] + c_4 \left[ \cos (- \frac x K) + i \sin (- \frac x K) \right]$$
 * (3.3)
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From the trigonometric identity, it is defined that $$\sin (- \theta) = - \sin (\theta)$$, $$\cos (- \theta) = \cos (\theta)$$, $$\sinh (- \theta) = - \sinh (\theta)$$, and $$\cosh (- \theta) = \cosh (\theta)$$. Apply the relations to the equation of $$X(x)$$.
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Then the equation can be simplified,
 * style="width:95%" | $$\displaystyle X(x) = c_1 \left[\cosh(\frac x K) + \sinh (\frac x K)\right] + c_2 \left[\cosh(\frac x K) - \sinh (\frac x K)\right]+ c_3 \left[ \cos (\frac x K) + i \sin (\frac x K) \right] + c_4 \left[ \cos (\frac x K) - i \sin (\frac x K) \right]$$
 * (3.4)
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 * style="width:95%" | $$\displaystyle X(x) = [c_1 + c_2] \cosh(\frac x K) + [c_1 - c_2] \sinh (\frac x K) + [c_3 + c_4] \cos (\frac x K) + [c_3 - c_4] i \sin (\frac x K)$$
 * (3.5)
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Author and proof-reader
[Author]
 * Jo

[proof-reader]
 * Raul Riveros

= References =

= Contributing members = Mike Faraone: Solved and Proofed Problems 1 & 2.

Jo: Authored Problem 3.

Rob Carroll: Solved and proof read problem 2

Michael Steinberg: Solved problem 2.

Raul Riveros: Proofed problems 2 and 3, edited formatting for all problems.