# symmetry methods for differential equations,exercise 1.4

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79 % ----------------------------------------------------------------------------------------
80 %    TITLE
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82 \begin{document}
83 \begin{CJK}{UTF8}{gkai}
84 \title{\textbf{Symmetry Methods for Differential Equations:\\Exercise 1.4}}
85 % \setlength\epigraphwidth{0.7\linewidth}
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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109 % ----------------------------------------------------------------------------------------
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(421)  评论(0编辑  收藏  举报