User:Egm6321.f12.team6.bang/report2

=Report 2=

Problem Statement

 * Consider the following function $$ \displaystyle \phi \left ( x, y \right ) = x^2y^\frac{3}{2} + \log(x^3y^2) = k $$
 * a) Find $$ \displaystyle G(y',y,x) = \frac{d}{dx}\phi(x,y)=0 $$
 * b) Show that (3) p.9-2 is an N1-ODE

Given

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$$ \displaystyle \phi \left ( x, y \right ) = x^2y^\frac{3}{2} + \log(x^3y^2) = k $$
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 * (5.1)
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Nomenclature

 * N1-ODE stands for Nonlinear First Order Differential Equation
 * $$ \displaystyle \frac{dy}{dx} = y' $$


 * $$ \displaystyle \frac{1}{x} = x^{-1} ; \displaystyle \frac{1}{y} = y^{-1} $$


 * $$ \displaystyle M_y(x,y) = \frac {\partial M(x,y)}{\partial y}; N_x(x,y) = \frac{\partial N(x,y)}{\partial x} $$

Part a): Derivation of G(y',y,x) Form

 * Using Chain Rule, we can take $$ \displaystyle \frac{d}{dx} \phi(x,y(x)) $$, which will becomes following.
 * $$ \displaystyle \frac{d\phi}{dx} = \frac{\partial \phi} {\partial x} \cancelto{1}\frac{dx}{dx} + \frac{\partial \phi} {\partial y} \frac{dy}{dx} = \frac{\partial \phi} {\partial x} + \frac{\partial \phi} {\partial y} \frac{dy}{dx} $$
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$$ \displaystyle \frac{d\phi}{dx} = \frac{\partial \phi} {\partial x} + \frac{\partial \phi} {\partial y} y' $$
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 * (5.2)
 * }
 * In order to substitute Equation 5.1 to Equation 5.2, take the derivative of $$ \displaystyle \phi \left ( x, y \right ) = x^2y^\frac{3}{2} + \log(x^3y^2) $$ respect to x and y.
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$$ \displaystyle \frac{\partial \phi} {\partial x} = 2xy^\frac{3}{2} + \frac {3x^2y^2} {x^3y^2} = 2xy^\frac{3}{2} + 3x^{-1} $$
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 * (5.3)
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$$ \displaystyle \frac{\partial \phi} {\partial y} = \frac{3}{2}xy^\frac{1}{2} + \frac {2x^3y} {x^3y^2} = \frac{3}{2}xy^\frac{1}{2} + 2y^{-1} $$
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 * (5.4)
 * }
 * Also k will become 0 since k is a constant value. Therefore the expression of $$ \displaystyle \frac{d\phi}{dx} $$ becomes following.
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$$ \displaystyle G(y',y,x) = \frac{d}{dx} \phi(x,y)= 2xy^\frac{3}{2} + 3x^{-1} + \left (\frac{3}{2}xy^\frac{1}{2} + 2y^{-1}\right )y' = 0$$
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 * (5.5)
 * }

Part b): Proof of N1-ODE

 * Exact N1-ODEs are subset of all sets of N1-ODEs. Also, in order to satify exactness condition for N1-ODEs, two conditions have to meet.
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1. $$ \displaystyle G(x,y,z) = M(x,y) + N(x,y)y' = 0$$
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 * (5.6)
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2. $$ \displaystyle M_y(x,y) = N_x(x,y) $$
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 * (5.7)
 * }
 * This means that if we prove that given equation is exact, then we can conclude that the given equation is N1-ODE. Using Equation 5.5 can prove the first condition that is written as Equation 5.6. This is demonstrated below.
 * $$ \displaystyle G(y',y,x) = M(x,y) + N(x,y)y' = \left (2xy^\frac{3}{2} + 3x^{-1}\right ) + \left (\frac{3}{2}x^2y^\frac{1}{2} + 2y^{-1}\right )y' = 0$$
 * In addition, we can identify both M and N.
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$$ \displaystyle M(x,y) = 2xy^\frac{3}{2} + 3x^{-1}$$
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 * (5.8)
 * }
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$$ \displaystyle N(x,y) = \frac{3}{2}x^2y^\frac{1}{2} + 2y^{-1}$$
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 * (5.9)
 * }
 * Then, we can test the second condition.
 * $$ \displaystyle \frac {\partial M(x,y)}{\partial y} = \frac{3}{2} \times 2xy^\frac{1}{2} = 3xy^\frac{1}{2} $$
 * $$ \displaystyle \frac {\partial N(x,y)}{\partial x} = 2 \times \frac{3}{2}x^2y^\frac{1}{2} = 3xy^\frac{1}{2} $$
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$$ \displaystyle M_y(x,y) = N_x(x,y) = 3xy^\frac{1}{2} $$
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 * (5.10)
 * }
 * As a result, since Equation 5.10 satifies the condition given in Equation 5.7, we can say that this equation is an Exact N1-ODE.
 * Furthermore, we can observe the linearity briefly using an answer from Part a). There are two criteria that a differential equation must meet in order to be linear.


 * 1) The dependnet variable y and all its derivatives are of the first degree; that is, the power of each term involving y is 1.
 * 2) Each coefficient depends on only the independent varible x.
 * The Equation 5.5 has coefficient depends on y, and also a y term does not have a power of 1. Therefore the equation is a nonlinear differential equation.
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Since we have tested that the given equation is nonlinear and exact, we can conclude that the given equation is Exact N1-ODE.
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