Answer
$c_1 = 1$
$c_2 = -2 $
$y = e^{-2x} -2 e^x$
Work Step by Step
$y = c_1e^{-2x} +c_2e^x$
$y' = -2c_1e^{-2x} +c_2e^x$
$y'' = 4c_1e^{-2x} +c_2e^x,$
Plug in the value $y(0) = -1$
$y(0) = c_1e^{-2(0)} +c_2e^(0) = -1$
$c_1 + c_2 = -1 ............... (1)$
Plug in the value $y'(0) = -4$
$y'(0) = -2c_1e^{-2(0)} +c_2e^0$
$-2c_1 + c_2 = -4..............(2)$
Solving equation (1) and (2) simultaneously, we get
$c_1 = 1$
$c_2 = -2 $
Rewrite the original equation with values of $c_1$ and $c_2$
$y = e^{-2x} -2 e^x$
