Answer
See below
Work Step by Step
Given: $y''+7y'+10y=0$
Substitute: $y''(x)+7y'(x)+10y(x)\\
=r^2e^{rx}+7re^{rx}+10e^{rx}\\
=e^{rx}(r^2+7r+10)\\
=0$
Since $e^{rx} \ne0 \rightarrow r^2+7r+10=0\\
\rightarrow (r+2)(r+5)=0\\
\rightarrow r=-2, r=-5$
The solutions to the given problem are: $y_1(x)=e^{-x}\\
y_2(x)=e^{-5x}$
Obtain: $W_{[y_1,y_2]}(x)=\begin{vmatrix}
e^{-2x} & e^{-5x}\\
-2e^{-2x}& -5e^{-5x}
\end{vmatrix}=-5^{-7x}-(-2e^{-7x})=-3e^{-7x}$
Since $-3e^{-7x} \ne0$, $y_1,y_2$ are linearly independent solutions to the equation.
Hence, the solutions is $y(x)=C_1e^{-2x}+C_2e^{-5x}$
