Question

Use the variation of parameters method to solve the DE y"+y'- 2y=1

Answer 1

`y(t)\ =\ C_1.e^(-2t)+C_2e^t-\ t.(e^(-2t))/(3)-(1)/(3)`

Explanation:

Given differential equation is,

y"+y'-2y=1

`=>\ (D^2+D-2D)y\ =\ 1`

To find the complementary function we will write,

`D^2+D-2=0`

`=>\ D\ =\ (-1+√(1^2+4* 2* 1))/(2* 1)\ or\ (-1-√(1^2+4* 2* 1))/(2* 1)`

`=>\ D\ =\ -2\ or\ 1`

Hence, the complementary function can be given by

`y(t)\ =\ C_1e^(-2t)\ +\ C_2e^t`

Let's say,

`y_1(t)\ =\ e^(-2t)\ \ =>y'_1(t)\ =\ -2e^(-2t)`

`y_2(t)\ =\ e^(t)\ \ =>y'_2(t)\ =\ e^(t)`

`g(t)\ =\ 1`

Wronskian can be given by,

`W\ =\ y_1(t).y'_2(t)\ -\ y_2(t).y'_1(t)`

`=\ e^(-2t).e^(t)\ -\ e^(t).(-2e^(-2t))`

`=\ e^(-t)\ +\ 2e^(-t)`

`=\ 3.e^(-t)`

Now, the particular integral can be given by

`y_p(t)=\ -y_1(t)\int(y_2(t).g(t))/(W)dt\ +\  y_2(t)\int(y_1(t).g(t))/(W)dt`

`=\ -e^(-2t)\int(e^t.1)/(3.e^(-t))+e^t\int(e^(-2t).1)/(3.e^(-t))dt`

`=\ -e^(-2t)\int(1)/(3)dt+(e^t)/(3)\int e^(-t)dt`

`=\ (-e^(-2t))/(3).t\ -\ (e^t)/(3).e^(-t)`

`=\ -t.(e^(-2t))/(3)-(1)/(3)`

Hence, the complete solution can be given by

`y(t)\ =\ C_1.e^(-2t)+C_2e^t-\ t.(e^(-2t))/(3)-(1)/(3)`