symmetry methods for differential equations,exercise 1.4

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 84 \title{\textbf{Symmetry Methods for Differential Equations:\\Exercise 1.4}} 
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 86 \author{\small{叶卢庆}\\{\small{杭州师范大学理学院,学号:1002011005}}\\{\small{Email:h5411167@gmail.com}}} % Institution
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110 %    ESSAY BODY
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112 \begin{exercise}[1.4]
113 Determine the value of $\alpha$ for which 
114 $$
115 (x',y')=(x+2\va,ye^{\alpha\va})
116 $$
117 is a symmetry of 
118 $$
119 \frac{dy}{dx}=y^2e^{-x}+y+e^x
120 $$
121 for all $\va\in\mathbf{R}$.  
122 \end{exercise}
123 \begin{proof}
124   The symmetry condition for the differential equation is 
125 $$
126 \frac{\frac{\pa g}{\pa x}+\frac{\pa g}{\pa y}w(x,y)}{\frac{\pa f}{\pa
127     x}+\frac{\pa f}{\pa y}w(x,y)}=w(f(x,y),g(x,y)).
128 $$
129 Where
130 $w(x,y)=y^2e^{-x}+y+e^x$,$f(x,y)=x+2\va,g(x,y)=ye^{\alpha\va}$.So the
131 symmetry condition can be written as 
132 $$
133 y^2e^{-x+\alpha\va}+e^{x+\alpha\va}=y^2e^{2\alpha\va-x-2\va}+e^{x+2\va}.
134 $$
135 So $\alpha=2$.
136 \end{proof}
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posted @ 2013-11-12 23:47  叶卢庆  阅读(349)  评论(0编辑  收藏