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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:}}\\{\small{Email:h5411167@gmail.com}}} % Institution

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 \begin{exercise}[1.4]

 Determine the value of $\alpha$ for which

 $$

 (x',y')=(x+\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

 $w(x,y)=y^2e^{-x}+y+e^x$,$f(x,y)=x+\va,g(x,y)=ye^{\alpha\va}$.So the

 symmetry condition can be written as

 $$

 y^2e^{-x+\alpha\va}+e^{x+\alpha\va}=y^2e^{\alpha\va-x-\va}+e^{x+\va}.

 $$

 So $\alpha=$.

 \end{proof}

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