Partial integral of d^2 y/dx^2-3dy/dx+2y=e^x

You need to solve the complementary equation `y'' - 3y' + 2y = 0`  to find the complementary solution, hence, supposing that one solution is proportional to `e^(tx),`  you need to substitute `e^(tx)`  for y in the given equation, such that:

`t^2*e^(tx) - 3t*e^(tx) + 2e^(tx) = 0`

You need to factor out `e^(tx)`  such that:

`e^(tx)(t^2 - 3t + 2) = 0`

Since `e^(tx)!=0` , hence only `t^2 - 3t + 2 = 0`  such that:

`t_(1,2) = (3+-sqrt(9 - 8))/2 => t_(1,2) = (3+-1)/2`  `t_1 = 2 ; t_2 = 1`

Hence, for`t = 1 => c_1e^(x) = y_1(x)`  and for `t = 2 =>c_2e^(2x) = y_2(x).`

Hence, evaluating the general solutions yields:

`y(x) = y_1(x) + y_2(x)`

`y(x) = c_1e^(x) + c_2e^(2x)`

You need to evaluate the particular solution such that:

`y_p = x*b*e^x => y'_p = b*e^x + b*e^x*x`

`y''_p = b*e^x*x + 2b*e^x`

`b(e^x*x + 2e^x) - 3(b*e^x + b*e^x*x) + 2be^x*x = e^x`

Equating the coefficients of `e^x`  yields `b = -1 => y_p = -x*e^x`

Hence, evaluating the general solution to the given equation yields `y = c_1e^(x) + c_2e^(2x) - x*e^x.`

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