Question

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

Answer 1

CF+PI=
`c_1e^(2x)+c_2e^(x)`+
`2e^(3t)`

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 =
`(4e^(3t))/((m-1)(m-2))`

at m= 3 in PI

`PI=(4e^(3t))/(2)=2e^(3t)`

so the complete solution is

CF+PI=
`c_1e^(2x)+c_2e^(x)`+
`2e^(3t)`