Question

у'' - 4у'+3у=0, у(0)=3, у' (0)=5

Answer 1

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