symmetry methods for differential equations,exercise 1.4

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\title{\textbf{Symmetry Methods for Differential Equations:\\Exercise 1.4}}
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\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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\begin{exercise}[1.4]
Determine the value of $\alpha$ for which
$$
(x',y')=(x+2\va,ye^{\alpha\va})
$$
is a symmetry of
$$
\frac{dy}{dx}=y^2e^{-x}+y+e^x
$$
for all $\va\in\mathbf{R}$.
\end{exercise}
\begin{proof}
The symmetry condition for the differential equation is
$$
\frac{\frac{\pa g}{\pa x}+\frac{\pa g}{\pa y}w(x,y)}{\frac{\pa f}{\pa
x}+\frac{\pa f}{\pa y}w(x,y)}=w(f(x,y),g(x,y)).
$$
Where
$w(x,y)=y^2e^{-x}+y+e^x$,$f(x,y)=x+2\va,g(x,y)=ye^{\alpha\va}$.So the
symmetry condition can be written as
$$
y^2e^{-x+\alpha\va}+e^{x+\alpha\va}=y^2e^{2\alpha\va-x-2\va}+e^{x+2\va}.
$$
So $\alpha=2$.
\end{proof}
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