User:Egm6321.f12.team6.bang/report3/problem*3.6

=Report 3= ==Problem *3.6: Validation of Exactness and Finding the Integrating Factor h ==

Problem Statement

 * Show that $$ \displaystyle \left( \frac{1}{3}x^3 + d_1 \right) \left(y^4 \right)y' + \left(5x^3 + 2 \right) \left(\frac{1}{5}y^5 + \sin{x} + d_2 \right) = 0 $$
 * a) either is exact, or
 * b) can be made exact by the IFM. Find the integrating factor h.

Given

 * 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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$$ \displaystyle \left( \frac{1}{3}x^3 + d_1 \right) \left(y^4 \right)y' + \left(5x^3 + 2 \right) \left(\frac{1}{5}y^5 + \sin{x} + d_2 \right) = 0 $$
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 * (6.1)
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$$ \displaystyle \frac {h_x}{h}=-\frac{1}{N}(N_x-M_y) $$
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 * (6.2)
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$$ \bar b(x,y) = \int^x b(s)ds + k_1(y) $$
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 * (6.3)
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Nomenclature

 * N1-ODE stands for Nonlinear First Order Differential Equation.
 * IFM stands for Euler Integrating Factor Method, or just simply Integrating Factor Method.
 * $$ \displaystyle y' = \frac{dy}{dx} $$
 * $$ \displaystyle M_y = \frac{\partial M(x,y)}{\partial y}; N_x = \frac{\partial N(x,y)}{\partial x} $$

Part a): Validation of Exactness

 * In order for a differential equation to be a exact, there are two conditions that must be met. Following are those two conditions:
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1. $$ \displaystyle M(x,y) + N(x,y)y' = 0 $$
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 * (6.4)
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2. $$ \displaystyle M_y(x,y) = N_x(x,y) $$
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 * (6.5)
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 * The first condition is examined. We can identify Equation 6.1 into following format.
 * $$ \displaystyle \underbrace{\left(5x^3 + 2 \right) \left(\frac{1}{5}y^5 + \sin{x} + d_2 \right)}_{M(x,y)} + \underbrace{\left( \frac{1}{3}x^3 + d_1 \right) \left(y^4 \right)}_{N(x,y)}y' = 0 $$
 * This can clearly express again as form of Equation 6.4.
 * Therefore, the first condition satisfies.
 * Now, we can proceed to the second condition. We identify both $$ \displaystyle M(x,y) $$, and $$ \displaystyle N(x,y) $$ from the first condition.
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$$ \displaystyle M(x,y) = \left(5x^3 + 2 \right) \left(\frac{1}{5}y^5 + \sin{x} + d_2 \right) $$
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 * (6.6)
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$$ \displaystyle N(x,y) = \left( \frac{1}{3}x^3 + d_1 \right) \left(y^4 \right) $$
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 * (6.7)
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 * Taking a partial derivative into $$ \displaystyle M_y(x,y) $$, and $$ \displaystyle N_x(x,y) $$ will give following equations.
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$$ \displaystyle M_y(x,y) = \left(5x^3 + 2 \right)\left(y^4 \right)$$
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 * (6.8)
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$$ \displaystyle N_x(x,y) = x^2y^4 $$
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 * (6.9)
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 * Comparing with both partial derivatives, we can conclude that $$ \displaystyle M_y(x,y) \neq N_x(x,y) $$.
 * Therefore, the second condition fails.
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Since the equation fails to satisfy one of conditions, this equation is not in a form of exact N1-ODE.
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Part b): Confirming the IFM, and Solving for Integrating Factor h

 * Assume that h is an integrating factor. Then we can rewrite Equation 6.2 into following format.
 * $$ \displaystyle \underbrace {(hM)}_{\bar {M}} + \underbrace {(hN)}_{\bar {N}} y' = 0 $$
 * If this equation is exact then the second condition has to meet. Thus applying the second condition for exactness is the following.
 * $$ \displaystyle {\bar {M}}_y = {\bar {N}}_x $$
 * Above equation is the same thing as the following equation.
 * $$ \displaystyle h_yM + hM_y = h_xN + hN_x $$ or rewriting as,
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$$ \displaystyle h_xN - h_yM + h \left(N_x-M_y \right) = 0 $$
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 * (6.10)
 * }
 * Then, the next step is to solve for h. The general class of N1-ODE can be express in the form of
 * $$ \displaystyle \bar b(x,y)c(y)y' + a(x) \bar c(x,y) = 0 $$.
 * This is indicating that $$ \displaystyle M(x,y) = a(x) \bar c(x,y) $$, and $$ \displaystyle N(x,y) = \bar b(x,y)c(y) $$.
 * Taking a partial derivative into $$ \displaystyle M_y $$, and $$ \displaystyle N_x $$ will provide in to this form.
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$$ \displaystyle M_y(x,y) = a(x)c(y) $$
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 * (6.11)
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$$ \displaystyle N_x(x,y) = b(x)c(y) $$
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 * (6.12)
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 * Also, we are given Equation 6.2. We can rewrite this equation using terms of $$ \displaystyle a(x) $$, $$ \displaystyle \bar b(x) $$, and $$ \displaystyle c(x) $$.
 * $$ \displaystyle \frac{h_x}{h} = -\frac{1}{N}(N_x-M_y) = -\frac{1}{\bar b(x)c(y)}[b(x)c(y)-a(x)c(y)] $$
 * Note that there is a given condition for an equation to render it exact. For more detail, please refer to Problem *3.5.
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$$ \displaystyle \frac{h_x}{h}= -\frac{1}{\bar b(x)}[b(x)-a(x)] $$
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 * (6.13)
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 * From Equation 6.6 and Equation 6.7, we can identify $$ \displaystyle a(x) $$, and $$ \displaystyle \bar b(x) $$. Furthermore, as Equation 6.3, $$ \displaystyle b(x) $$ is simply taking the derivative of $$ \displaystyle \bar b(x) $$.
 * $$ \displaystyle a(x) = 5x^3 + 2 $$
 * $$ \displaystyle \bar b(x) = \frac {1}{3}x^3+d_1 $$
 * $$ \displaystyle b(x) = x^2 $$
 * Solving for $$ \displaystyle \frac{h_x}{h} $$, when $$ \displaystyle d_1 = 0. $$
 * $$ \displaystyle \frac{h_x}{h} = -\frac{1}{\bar b(x)}[b(x)-a(x)] $$
 * $$ \displaystyle = -\frac{1}{\frac {1}{3}x^3}[x^2 - (5x^3 + 2)] = -\frac {3}{x^3}[x^2 - 5x^3 - 2)] $$
 * $$ \displaystyle = 15 - 3x^{-1} + 6x^{-3} $$
 * $$ \displaystyle \int \frac{h_x}{h} = \int^x (15 - 3s^{-1} + 6s^{-3})ds + k $$
 * $$ \displaystyle \ln{h(x)} = 15x - 3\ln x - 3x^{-2} + k $$
 * $$ \displaystyle h = e^{15x - 3\ln x - 3x^{-2} + k} = e^{15x - 3x^{-2} + k} \cdot e^{-3\ln x} = e^{15x - 3x^{-2} + k} \cdot e^{\ln {x^-3}} $$


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 * $$ \displaystyle = x^{-3}e^{15x - 3x^{-2} + k} $$


 * (6.14)
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 * Finally we recognize that the integration constant $$ k $$  is not necessary, and therefore eliminate it from the integration factor. For more detail on why this is possible, please refer to Problem *3.1.


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Therefore, this equation can be made exact by the IFM. The integrating factor of h is $$ \displaystyle x^{-3}e^{15x - 3x^{-2}} $$.
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