У'' - 4у'+3у=0, у(0)=3, у' (0)=5

Решите задачу:

$y''-4y'+3y=0 \\ y=e^{kx} \\ k^2e^{kx}-4ke^{kx}+3e^{kx}=0 \\ e^{kx}(k^2-4k+3)=0 \\ k^2-4k+3=0 \\ (k-3)(k-1)=0 \\ y=C_1e^{3x}+C_2e^x \\ \left \{ {{C_1e^{0}+C_2e^0=3} \atop {3C_1e^{0}+C_2e^0=5}} \right. \\ \left \{ {{C_1+C_2=3} \atop {3C_1+C_2=5}} \right. \\ \left \{ {{C_2=3-C_1} \atop {3C_1+3-C_1=5}} \right. \\ \left \{ {{C_2=3-C_1} \atop {2C_1=2}} \right. \\ \left \{ {{C_2=3-1=2} \atop {C_1=1}} \right. \\\\ Y=e^{3x}+2e^x$