User:Egm6321.f12.team6.bang/report3/problem*3.7

=Report 3= ==Problem *3.7: Formulating N1-ODE (Exact, or Exact using IFM) Equation ==

Problem Statement

 * $$ \displaystyle a(x) = \sin {x^3} $$
 * $$ \displaystyle b(x) = \cos {x} $$
 * $$ \displaystyle c(y) = \exp {2y} $$
 * 1. Find an N1-ODE of the $$ \displaystyle \underbrace{\bar b(x,y)c(y)}_{N(x,y)}y' + \underbrace{a(x)\bar c(x,y)}_{M(x,y)} = 0 $$ that is either exact or can be made exact by IFM.
 * 2. Find the first integral $$ \displaystyle \phi(x,y) = k $$

Given

 * $$ \displaystyle a(x) = \sin {x^3} $$
 * $$ \displaystyle b(x) = \cos {x} $$
 * $$ \displaystyle c(y) = \exp {2y} $$
 * Problem *3.5 proved that an integrating factor h(x) can be found to render it exact, only if $$ \displaystyle k_1(y)=d_1 $$ (constant).
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$$ \bar b(x,y) = \int^x b(s)ds + k_1(y) $$
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 * (7.1)
 * }
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$$ \bar c(x,y) = \int^y c(s)ds + k_2(x) $$
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 * (7.2)
 * }

Nomenclature

 * N1-ODE stands for Nonlinear First Order Differential Equation.
 * IFM stands for Euler Integrating Factor Method, or just simply Integrating Factor Method.
 * $$\displaystyle \exp(2y)=e^{2y} $$

Part 1: Forming N1-ODE Equation

 * The general form of N1-ODE is written as following
 * $$ \displaystyle \bar b(x,y)c(y)y' + a(x)\bar c(x,y) = 0 $$
 * However, Problem 3.5 proved that $$ \displaystyle k_1(y) $$ has to be equal to constant. Therefore, we can rewrite general class of N1-ODE that are either exact, or can be made exact by the IFM has the form of following.
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$$ \displaystyle \bar b(x)c(y)y' + a(x)\bar c(x,y) = 0 $$
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 * (7.3)
 * }
 * In order to form such an equation, we need $$ \bar b(x)$$, and $$ \bar c(x)$$. Using Equation 7.1, and Equation 7.2, we can derive both terms.
 * $$ \displaystyle \bar b(x) = \int^x b(s)ds + k_1 = \int^x \cos(s)ds + k_1 $$
 * $$ \displaystyle = \sin(x) + k_1 $$
 * $$ \displaystyle \bar c(x,y) = \int^y c(s)ds + k_2(x) = \int^y e^{2s}ds + k_2(x) $$
 * $$ \displaystyle = \frac{e^{2y}}{2}+k_2(x) + c $$
 * Just for simplicity, set values $$ \displaystyle k_1=k_2=c=0 $$.
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$$ \displaystyle \bar b(x) = \sin(x) $$
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 * (7.4)
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$$ \displaystyle \bar c(x,y) = \frac{e^{2y}}{2} $$
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 * (7.5)
 * }
 * Then we have all components to form N1-ODE. Plug each terms into Equation 7.3 will give the following.
 * $$ \displaystyle \sin(x)e^{2y}y' + \sin{x^3}\frac{e^{2y}}{2} = 0 $$
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Therefore N1-ODE becomes $$ \displaystyle \sin x \cdot y' + \frac{1}{2} \sin{x^3}= 0 $$
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 * (7.6)
 * }

Part 2: First Integral of Ø(x,y)

 * From the previous part, we derive the N1-ODE using the condition of exactness. Refering to Section 8-6 example, we can reverse engineer to find $$ \displaystyle \phi(x,y)=k $$. We know that first total derivative is written in the following form.
 * $$ \displaystyle \frac{d \phi}{dx} = \frac {\partial \phi}{\partial x} + \frac {\partial \phi}{\partial y}\frac {dy}{dx} = 0 $$
 * However, referring to the Section 8-6 Example, taking the integral of $$ \displaystyle \frac {d\phi}{dx} $$ will simply give the expression in terms of $$ \displaystyle \phi $$. Due to the derivative, the 0 of righthand side will have a constant. The equation becomes the following.
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$$ \displaystyle \phi (x,y)= \int \phi_x dx + \int \phi_y dy = c_1 $$
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 * (7.7)
 * }
 * Also, note that three following four terms are the same.
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$$ \displaystyle \frac {\partial \phi}{\partial x} = \phi_x(x,y) = M(x,y) = \frac {\sin x^3}{2} $$
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 * (7.8)
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$$ \displaystyle \frac {\partial \phi}{\partial y} = \phi_y(x,y) = N(x,y) = \sin x $$
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 * (7.9)
 * }
 * Taking the integral both Equation 7.8, and Equation 7.9
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$$ \displaystyle \int \phi_x dx = \int \frac {\sin x^3}{2} dx = -\frac {\cos x^3}{4x^2} + c_2 $$
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 * (7.10)
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$$ \displaystyle \int \phi_y dy = \int \sin x dy = y \sin x + c_3 $$
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 * (7.11)
 * }
 * Put Equation 7.10, and Equation 7.11 into Equation 7.7.
 * $$ \displaystyle \phi (x,y)= -\frac {\cos x^3}{4x^2} + c_2 + y \sin x + c_3 = c_1 $$
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$$ \displaystyle \phi (x,y)= y \sin x - \frac {\cos x^3}{4x^2} = c_1 + c_2 + c_3 $$
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 * (7.12)
 * }
 * We can combine all constant terms of c1, c2, and c3 into k. Finally equation can be expressed with k constant.
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$$ \displaystyle \phi (x,y)= y \sin x - \frac {\cos x^3}{4x^2} = k $$
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 * (7.13)
 * }