## Calculus, 10th Edition (Anton)

$$y = A{e^{2x}} + B{e^{ - 4x}}{\text{ satisfies the equation}}$$
\eqalign{ & y = A{e^{2x}} + B{e^{ - 4x}} \cr & {\text{Find }}y'{\text{ and }}y'' \cr & y' = \left( {A{e^{2x}}} \right)' + \left( {B{e^{ - 4x}}} \right)' \cr & y' = \left( {2A{e^{2x}}} \right) + \left( { - 4B{e^{ - 4x}}} \right) \cr & y' = 2A{e^{2x}} - 4B{e^{ - 4x}} \cr & \cr & y'' = \left( {2A{e^{2x}}} \right)' + \left( { - 4B{e^{ - 4x}}} \right)' \cr & {\text{pull out constants}} \cr & y'' = 2A\left( {{e^{2x}}} \right)' - 4B\left( {{e^{ - 4x}}} \right)' \cr & y'' = 2A\left( {2{e^{2x}}} \right) - 4B\left( { - 4{e^{ - 4x}}} \right) \cr & y'' = 4A{e^{2x}} + 16B{e^{ - 4x}} \cr & \cr & {\text{substituting }}y'{\text{ and }}y''{\text{ into the equation }}y'' + 2y' - 8y = 0 \cr & 4A{e^{2x}} + 16B{e^{ - 4x}} + 2\left( {2A{e^{2x}} - 4B{e^{ - 4x}}} \right) - 8\left( {A{e^{2x}} + B{e^{ - 4x}}} \right) = 0 \cr & 4A{e^{2x}} + 16B{e^{ - 4x}} + 4A{e^{2x}} - 8B{e^{ - 4x}} - 8A{e^{2x}} - 8B{e^{ - 4x}} = 0 \cr & {\text{reduce terms}} \cr & \left( {4A{e^{2x}} + 4A{e^{2x}} - 8A{e^{2x}}} \right) + \left( {16B{e^{ - 4x}} - 8B{e^{ - 4x}} - 8B{e^{ - 4x}}} \right) = 0 \cr & 0 = 0 \cr & {\text{Thus}}{\text{, }}y = A{e^{2x}} + B{e^{ - 4x}}{\text{ satisfies the equation}} \cr}