Question

y''+y'-6y=0 если y=0 и y'=-10 при х=0

Answer 1

$y''+y'-6y=0$
$y=e^{kx}$
$(e^{kx})''+(e^{kx})'-6(e^{kx})=0$
$k^2e^{kx}+ke^{kx}-6e^{kx}=0$
$e^{kx}(k^2+k-6)=0$
$k^2+k-6=0$
$(k+3)(k-2)=0$
$k_1=-3; k_2=2$
$Y=C_1e^{-3x}+C_2e^{2x}$

$y(0)=0;y'(0)=-10$
$left { {{C_1+C_2=0} atop {-3C_1+2C_2=-10}} right.$
$left { {{C_1=-C_2} atop {-3C_1+2C_2=-10}} right.$
$left { {{C_1=-C_2} atop {3C_2+2C_2=-10}} right.$
$left { {{C_1=-C_2} atop {5C_2=-10}} right.$
$left { {{C_1=2} atop {C_2=-2}} right.$

$Y=2e^{-3x}-2e^{2x}$