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\item[1.]$\bold{Motivation}$ From $e^x$, we find $(e^x)' = e^x$. Immediately the question arises: Is $e^x$ unique for $f'(x) = f(x)$? How about the case $f'(x) = \alpha f(x)$?
\item[2.]$\bold{Solution}$ By Taylor Expansion, we find analytic solutions $c_0\cdot e^{\alpha x}$ and by assumption that $f(x)\textgreater 0$ we get the same solutions. 
\item[3.]$\bold{Generalization}$ We then hope to solve $y' + p(x)y = q(x) $. We perform multiplication of the function $\varphi (x)$, getting its general form. Substituting $-\alpha$ and $0$ for $p(x)$ and $q(x)$ resp. , we entirely solve (1).
\item[4.]$\bold{Second-ordered}$ The  equation $y" + (\alpha + \beta) y' + \alpha \beta\cdot y= 0$  can be  solved by means of two-variable simultaneous equations, and that of solving first-order equations twice. 
\item[5.]$\bold{Multiple-Root}$ Solution to $y" + 2r\cdot y' + r^2 y= 0$ is $y = c_0\cdot e^{-rx} + c_1\cdot xe^{-rx}$. 
\item[6.]$\bold{Imaginary-Roots}$ Two particular roots $y_1 = \sin ax$, $y_2=\cos ax$ would lead to the general solution $y = c_1 \sin ax + c_2 \cos ax$.
\item[7.]$\bold{Generalization}$ By use of "translation" like that performed on graphs on $\mathbb{R}^2$, the form is $y = e^{\rho x} (c_1 \sin \omega x + c_2 \cos \omega x)$. Also, we might apply theories of complex variables, and imitate previous methods. The solution set remains the same. 
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