User talk:EGM6321.F12.team5.nguyen.

R*3.6 Showing a function is exact or can be made exact

Given

Consider the following function


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$$\underbrace{\left(\frac{1}{3}x^3+d_1 \right )(y^4)}_{\displaystyle \color{blue}{N(x,y)}}\,y' + \underbrace{(5x^3+2)\left(\frac{1}{5}y^5+\sin x + d_2 \right)}_{\displaystyle \color{blue}{M(x,y)}}=0$$ (1)
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This problem was restated from sec. 13:

Find

Show that (1) either is exact, or ican be made exact by the IFM. Find the integrating factor $$ h $$.

Solution This solution was solved on our own.

Showing the Equation is Exact
In order for a function to be exact, it must satisfy two exactness conditions. 1. The equation can be written as
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$$M(x,y) + N(x,y)y' = 0$$ (3.6.1) From inspection of (1), we can conclude this condition is met where
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$$M(x,y) = (5x^3+2)(\frac{1}{5}y^5+\sin x + d_2)$$ $$N(x,y) = (\frac{1}{3}x^3+d_1)(y^4)$$ (3.6.2) 2. The equation meets the Second Exactness Condition:
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$$M_y(x,y)=N_x(x,y)$$ $$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$ (3.6.3)
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Get the Derivative of $$M(x,y)$$ with respect to $$y$$
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$$\frac{\partial M(x,y)}{\partial y}=[(5x^3+2)(\frac{1}{5}y^5+\sin x + d_2)]\frac{d}{dy}$$ (3.6.4)
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By applying the Product Rule $$(uv)' = u'v+uv'$$, the equation can be re-written
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$$\frac{\partial M(x,y)}{\partial y}=(5x^3+2)\frac{d}{dy}(\frac{1}{5}y^5+\sin x + d_2)+(5x^3+2)(\frac{1}{5}y^5+\sin x + d_2)\frac{d}{dy}$$ (3.6.5)
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$$\frac{\partial M(x,y)}{\partial y}=\cancelto{0}{(5x^3+2)}(\frac{1}{5}y^5+\sin x + d_2)+(5x^3+2)(5*\frac{1}{5}y^4+\cancelto{0}{\sin x} + \cancelto{0}{d_2})$$ (3.6.6)
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$$\frac{\partial M(x,y)}{\partial y}=(5x^3+2)(y^4)$$ (3.6.7)
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Get the Derivative of $$N(x,y)$$ with respect to $$x$$
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$$\frac{\partial N(x,y)}{\partial x}=[(\frac{1}{3}x^3+d_1)(y^4)]\frac{d}{dx}$$ (3.6.8)
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By applying the Product Rule $$(uv)' = u'v+uv'$$, the equation can be re-written
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$$\frac{\partial N(x,y)}{\partial x}=(\frac{1}{3}x^3+d_1)\frac{d}{dx}(y^4)+(\frac{1}{3}x^3+d_1)(y^4)\frac{d}{dx}$$ (3.6.9)
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$$\frac{\partial N(x,y)}{\partial x}=(3*\frac{1}{3}x^2+\cancelto{0}{d_1})(y^4)+(\frac{1}{3}x^3+d_1)\cancelto{0}{(y^4)}$$ (3.6.10)
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$$\frac{\partial N(x,y)}{\partial x}=(x^2)(y^4)$$ (3.6.10)
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Evaluating the Second Exactness Condition shows that the equation is not exact.
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$$\frac{\partial M(x,y)}{\partial y}=\frac{\partial N(x,y)}{\partial x}$$
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$$(5x^3+2)(y^4)\neq(x^2)(y^4)$$ (3.6.11) Since (1) is not exact, we can show that the equation can be made exact by the Integrating Factor Method.
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Integrating Factor Method
From sec. 11:, supposing $$ h_y(x,y)=0$$, the integrating factor is defined as
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$$h(x)=exp\left[\int^x n(s)ds+k \right ]$$ 11-2 (4)
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where
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$$\frac{h_x}{h}=- \frac{1}{N}(N_x-M_y)=: n(x)$$ 11-2 (2)
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Since (1) is a class of N1-ODE of the form:
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$$\underbrace{\bar b(x,y)c(y)}_{\displaystyle \color{blue}{N(x,y)}}\,y' + \underbrace{a(x)\bar c(x,y)}_{\displaystyle \color{blue}{M(x,y)}}=0$$ 13-2 (1)
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$$\underbrace{\left(\frac{1}{3}x^3+d_1 \right )(y^4)}_{\displaystyle \color{blue}{N(x,y)}}\,y' + \underbrace{(5x^3+2)\left(\frac{1}{5}y^5+\sin x + d_2 \right)}_{\displaystyle \color{blue}{M(x,y)}}=0$$ (1)
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Then we can re-define 11-2 (2) as
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$$\frac{h_x}{h}=- \frac{1}{N}(N_x-M_y)= \underbrace {-\frac{1}{\bar b(x) \cancel {c(y)}} \left[b(x) \cancel{c(y)}-a(x) \cancel {c(y)} \right ]}_{\displaystyle \color{blue}{n(x)}}$$ 3.6.12
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Where
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$$\bar b(x)= \frac{1}{3}x^3+d_1$$ 3.6.13
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$$b(x)= 3*\frac{1}{3}x^2+0*d_1=x^2$$ 3.6.14
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$$c(y)=y^4$$ 3.6.15
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$$a(x)=5x^3+2$$ 3.6.16
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$$\bar c(x,y)=\frac{1}{5}y^5+sinx+d_2$$ 3.6.17
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Substituting equations 3.6.13-3.5.17 back into 3.6.12, we get $$n(x)$$ which we can use to find the Integrating Factor $$ h(x) $$. Note: since $$d_1$$ will eventually go to zero once we take it's derivative, we will set $$d_1=0$$ in the remaining equations
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$$n(x) = - \frac{1}{\frac{1}{3}x^2}\left[x^2-5x^3-2 \right ]$$ 3.6.18
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$$h(x)=exp\left[\int^x n(s)ds+k \right ]$$ 11-2 (4)
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$$h(x)=exp\left[\int^x - \frac{1}{\frac{1}{3}s^2}\left[s^2-5s^3-2 \right ]ds+k \right ]$$ 3.6.19
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$$h(x)=exp\left[\int^x - \frac{3}{s^2}\left[s^2-5s^3-2 \right ]ds+k \right ]$$ 3.6.19
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$$h(x)=exp\left[\int^x (-3+15s-2s^2)ds+k\right ]$$ 3.6.20
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$$h(x)=exp\left[\int^x (\cancel{-3}+15-4s)+k\right ]$$ 3.6.21
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$$h(x)=exp\left[15-4x+k\right]$$ 3.6.22
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= Old stuff =

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MaintenanceBot (discuss • contribs) 03:45, 11 November 2013 (UTC)