User:Egm6321.f2010.team5.riveros/hw4p1

= 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 $$y(x)$$

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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 * style="padding:10px; border:2px solid #8888aa" | (1.1) satisfies the 1st Condition of Exactness
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2nd Condition of Exactness

 * $$\displaystyle f \ = \ cos \ x $$
 * $$\displaystyle f_{xx} \ = \ -cos \ x $$
 * $$\displaystyle f_{xy} \ = \ f_{yy} \ = \ f_{xp} \ = \ f_{yp} \ = \ f_{y} \ = \ 0 $$
 * $$\displaystyle g \ = \ (x^2 - sin \ x) p + 2xy $$
 * $$\displaystyle g_{xp} \ = \ 2x - cos \ x $$
 * $$\displaystyle g_{y} \ = \ 2x $$
 * $$\displaystyle g_{yp} \ = \ g_{pp} \ = \ 0 $$


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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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from Mtg 10 (1) and (6) p.10-3


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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