Answer
$c_1 =-1$
$c_2 = -1$
$y = -e^{-2x}( sin 3x+cos 3x)$
Work Step by Step
$y = e^{-2x}(c_1 sin 3x+c_2 cos 3x) $
$y' = e^{-2x}[-(2c_1+3c_2) sin 3x+(3_c1-2c_2) cos 3x] $
Plug in the value $y(0) = -1$
$y(0) = e^{-2(0)}(c_1 sin 3(0)+c_2 cos 3(0))= -1$
$c_2 = -1$
Plug in the value $y'(0) = -1$
$y'(0) =e^{0}[-(2c_1+3c_2) sin 0+(3c_1-2c_2) cos 0] = -1$
$3c_1 - 2c_2= -1$
Where $c_2 = -1$
$3c_1 - 2(-1)= -1$
$c_1 = -1$
we get
$c_1 =-1$
$c_2 = -1$
Rewrite the original equation with values of $c_1$ and $c_2$
$y = -e^{-2x}( sin 3x+cos 3x)$
