Answer
See below
Work Step by Step
Given: $y''+4y'=0$
Substitute: $y''(x)+4y'(x)\\
=r^2e^{rx}+4re^{rx}\\
=e^{rx}(r^2+4r)\\
=0$
Since $e^{rx} \ne0 \rightarrow r^2+4r=0\\
\rightarrow r(r+4)=0\\
\rightarrow r=0, r=-4$
The solutions to the given problem are: $y_1(x)=1 \\
y_2(x)=e^{-4x}$
Obtain: $W_{[y_1,y_2]}(x)=\begin{vmatrix}
1 & e^{-4x}\\
0 & -4e^{-4x}
\end{vmatrix}=-4e^{-4x}$
Since $-4e^{-4x} \ne0$, $y_1,y_2$ are linearly independent solutions to the equation.
Hence, the solutions is $y(x)=C_1+C_2e^{-4x}$
