Question

Solve the Method of variation of Parameters. y" - 3y' + 2y = 4e^3t

Answer 1

Answer:

CF+PI=$c_1e^{2x}+c_2e^{x}$+$2e^{3t}$

Step-by-step explanation:

we have given y"-3y'=2y=$4e^{3t}$

this differential equation solution have two part that CF and PI

CALCULATION OF CF :

$m^2-3m+2=0$

$m^2-2m-m+2=0$

$(m-1)(m-2)=0$

m=1 and m=2

so CF=$c_1e^{2x}+c_2e^{x}$

CALCULATION OF PI :

PI =   $\frac{4e^{3t}}{(m-1)(m-2)}$

at m= 3 in PI

$PI=\frac{4e^{3t}}{2}=2e^{3t}$

so the complete solution is

CF+PI=$c_1e^{2x}+c_2e^{x}$+$2e^{3t}$