Answer:
CF+PI=[tex]c_1e^{2x}+c_2e^{x}[/tex]+[tex]2e^{3t}[/tex]
Step-by-step explanation:
we have given y"-3y'=2y=[tex]4e^{3t}[/tex]
this differential equation solution have two part that CF and PI
CALCULATION OF CF :
[tex]m^2-3m+2=0[/tex]
[tex]m^2-2m-m+2=0[/tex]
[tex](m-1)(m-2)=0[/tex]
m=1 and m=2
so CF=[tex]c_1e^{2x}+c_2e^{x}[/tex]
CALCULATION OF PI :
PI = [tex]\frac{4e^{3t}}{(m-1)(m-2)}[/tex]
at m= 3 in PI
[tex]PI=\frac{4e^{3t}}{2}=2e^{3t}[/tex]
so the complete solution is
CF+PI=[tex]c_1e^{2x}+c_2e^{x}[/tex]+[tex]2e^{3t}[/tex]
