Answer
$c_1 = -1$
$c_2 = 2 $
$y = -e^{-2x} + 2e^{3x} $
Work Step by Step
$y = c_1e^{-2x} + c_2e^{3x} $
$y' = -2c_1e^{-2x} + 3c_2e^{3x}$
Plug in the value $y(0) = 1$
$y(0) = c_1e^{-2(0)} + c_2e^{3(0)} = 1$
$c_1 + c_2 = 1 ............... (1)$
Plug in the value $y'(0) = 8$
$y'(0) = -2c_1e^{-2(0)} + 3c_2e^{3(0)}= 8$
$-2c_1 + 3c_2 = 8..............(2)$
Solving equation (1) and (2) simultaneously, we get
$c_1 = -1$
$c_2 = 2 $
Rewrite the original equation with values of $c_1$ and $c_2$
$y = -e^{-2x} + 2e^{3x} $
