User talk:EGM6321.F12.team5.nguyen R2-5

R2.5 Showing a function is a N1-ODE

Given

Consider the following function


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$$\phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3y^2)=k$$
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Find

Find
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$$G(y',y,x)=\frac{d}{dx}\phi (x,y)=0$$ and show that (3) p.9-2 is an N1-ODE.
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Solution This solution was solved on our own.

Find Total Derivative
First set the function equal to zero
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$$\phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3y^2)=k$$ (2.5.1)
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$$\phi(x,y)=x^2y^{\frac{3}{2}}+log(x^3y^2)-k=0$$ (2.5.2)
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To make differentiating simpler, use the following log law to separate out the log function into two separate log functions.
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$$ log(ab) = log(a)+log(b)$$
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$$\phi(x,y)=(x^2y^{\frac{3}{2}})+log(x^3)+log(y^2)-k$$
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The total derivative of a function can be defined as follows
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$$\frac{d\phi (x,y(x))}{dx}=\frac{\partial\phi(x,y(x)))}{\partial x}+\frac{\partial\phi(x,y(x)))}{\partial y}\frac{dy}{dx}$$
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Take the derivative of $$\phi(x,y)$$ with respect to $$x$$.
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$$\frac{\partial\phi(x,y(x))}{\partial x}=\frac{d}{dx}(x^2y^{\frac{3}{2}}+log(y^2)-k)$$ (2.5.3)
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$$\frac{\partial\phi(x,y(x))}{\partial x}=\frac{d}{dx}x^2y^{\frac{3}{2}}+\frac{d}{dx}log(x^3)+\cancelto{0}{\frac{d}{dx}log(y^2)}-\cancelto{0}{\frac{d}{dx}k)}$$ (2.5.4)
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$$\frac{\partial\phi(x,y(x))}{\partial x}=2xy^{\frac{3}{2}}+\frac{3}{\ln x}$$ (2.5.5)
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Take the derivative of $$\phi(x,y)$$ with respect to $$y$$.
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$$\frac{\partial\phi(x,y(x))}{\partial y}=\frac{d}{dy}(x^2y^{\frac{3}{2}}+log(x^3)+log(y^2)-k)$$ (2.5.6)
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$$\frac{\partial\phi(x,y(x))}{\partial y}=\frac{d}{dy}x^2y^{\frac{3}{2}}+\cancelto{0}{\frac{d}{dy}log(x^3)}+\frac{d}{dy}log(y^2)-\cancelto{0}{\frac{d}{dy}k)}$$ (2.5.7)
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$$\frac{\partial\phi(x,y(x))}{\partial y}=\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y}$$ (2.5.8)
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Therefore
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$$\frac{d\phi (x,y(x))}{dx}=\frac{\partial\phi(x,y(x)))}{\partial x}+\frac{\partial\phi(x,y(x)))}{\partial y}\frac{dy}{dx}$$
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$$\frac{d\phi (x,y)}{dx}=(2xy^{\frac{3}{2}}+\frac{3}{\ln x})+(\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y})\frac{dy}{dx}$$ (2.5.9)
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The solution can be written as follows in the particular form
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(2.5.10)
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$$G(y',y,x)= \underbrace{2xy^{\frac{3}{2}}+\frac{3}{\ln x}}_{M(x,y)}+\underbrace{\frac{3}{2}x^2y^{\frac{1}{2}})+\frac{2}{\ln y}}_{N(x,y)}\underbrace{\frac{dy}{dx}}_{y'}=0$$ (2.5.10)
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Show (3) 9-2 is an N1-ODE
As shown in (2.5.10),  $$G(y',y,x)$$  can be expressed in the particular form
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$$G(y',y,x)= M(x,y)+N(x,y)y'=0$$
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From inspection, it can be concluded that the equation is first-order.

To show that $$G(y',y,x)$$ is non-linear in  $$y'$$, it can be shown that
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$$G(\alpha u+\beta v, y, x) \neq \alpha G(u,y,x)+\beta G(v,y,x)$$
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From (2.5.10)
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$$G(y',y,x)= 2xy^{\frac{3}{2}}+\frac{3}{\ln x}+\frac{3}{2}x^2y^{\frac{1}{2}})+\frac{2}{\ln y}\frac{dy}{dx}=0$$
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$$G(\alpha u+\beta v,y,x)= 2xy^{\frac{3}{2}}+\frac{3}{\ln x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y}(\alpha +\beta )=0$$ (2.5.11)
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$$G(\alpha u+\beta v,y,x)= 2xy^{\frac{3}{2}}+\frac{3}{\ln x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y}\alpha +\frac{2}{\ln y}\beta =0$$ (2.5.12)
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From (2.5.8)
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$$\frac{\partial\phi(x,y(x))}{\partial y}=\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y}$$
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$$\alpha G(u,y,x)=\alpha (\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y})$$ (2.5.13)
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$$\alpha G(u,y,x)=\alpha \frac{3}{2}x^2y^{\frac{1}{2}}+\alpha \frac{2}{\ln y}$$ (2.5.14)
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From (2.5.5)
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$$\frac{\partial\phi(x,y(x))}{\partial x}=2xy^{\frac{3}{2}}+\frac{3}{\ln x}$$
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$$\beta G(v,y,x)=\beta (2xy^{\frac{3}{2}}+\frac{3}{\ln x})$$ (2.5.15)
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$$\beta G(v,y,x)=\beta 2xy^{\frac{3}{2}}+\beta \frac{3}{\ln x}$$ (2.5.16)
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Therefore, $$G(y',y,x)$$  is non-linear
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$$G(\alpha u+\beta v, y, x) \neq \alpha G(u,y,x)+\beta G(v,y,x)$$
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$$2xy^{\frac{3}{2}}+\frac{3}{\ln x}+\frac{3}{2}x^2y^{\frac{1}{2}}+\frac{2}{\ln y}\alpha +\frac{2}{\ln y}\beta \neq \alpha \frac{3}{2}x^2y^{\frac{1}{2}}+\alpha \frac{2}{\ln y} + \beta 2xy^{\frac{3}{2}}+\beta \frac{3}{\ln x} $$ (2.5.17) Conclusion: Equation (3) 9-2 is an N1-ODE
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