Question

For the nonhomogeneous equation, using the method of undetermined coefficients, the solution I got for `(d^2y)/(dx^2) - 5(dy)/(dx) + 6y = xe^x` is `y = (xe^x)/2 + (3e^x)/4`, but along the way I got two solutions for `A_1`?

NOTE: As usual, I figured it out a few minutes after I posted this. Oh well! Leaving it up for other people to see, then. :)

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I was letting `y = x(A_1x + A_0)e^x`, and I got `(dy)/(dx) = [A_1x^2 + (2A_1 + A_0)x + A_0]e^x` and `(d^2y)/(dx^2) = [A_1x^2 + (4A_1 + A_0)x + 2A_1 + 2A_0]e^x`. When I solved for `A_0` and `A_1` however, I got:

`x^2` terms:

`A_1 - 5A_1 + 6A_1 = 0 => A_1 = 0` is one solution to the quadratic (whoops, just realized this)

`x` terms:

`4A_1 + A_0 - 10A_1 - 5A_0 + 6A_0 = 1`, and if `A_1 = 0`, then:

`A_0 = 1/2`

Constant terms:

`2A_1 + 2A_0 - 5A_0 = 0`, and if `A_1 ne 0`, then `A_1 = 3/2A_0 = 3/4`. So the result is indeed `y = (xe^x)/2 + (3e^x)/4`.

Answer 1

`(d^2y)/(dx^2) - 5(dy)/(dx) + 6y = xe^x` is a linear non-homogeneous differential equation.

The solution `y_p = (xe^x)/2 + (3e^x)/4` is a particular solution for the differential equation.

The complete solution can be composed as the addition of a particular solution plus the homogeneous solution. The homogeneous solution has the structure

`y_h=C_1e^(2x)+C_2e^(3x)`

as can be easily verified, so the complete solution is

`y = y_h+y_p = C_1e^(2x)+C_2e^(3x)+ (xe^x)/2 + (3e^x)/4` without ambiguity in the `C_1,C_2` determination.

Answer 2

I'll use `A=A_0` and `B=A_1` to save typing

The problem is you are trying the wrong solution you have
`g(x)=(ax^2+b)e^x` and you should be trying `g(x)=(ax+b)e^x`

So for the PI we have:

`y=(Ax + B)e^x`

`y'=(Ax + B)(e^x) + (A)(e^x)`
`\ \ \ \=(Ax + B)e^x + Ae^x`

`y'' = (Ax + B)e^x + Ae^x+Ae^x`
`\ \ \ \ \ = (Ax + B)e^x + 2Ae^x`

Subs into the DE and we get:

`y''-5y'+6y=xe^x`

`:. {(Ax + B)e^x + 2Ae^x} -5{(Ax + B)e^x + Ae^x}+6{(Ax + B)e^x}=xe^x`

Compare coefficients of `xe^x`:

`A-5A+6A=1=>2A=1=>A=1/2`

Compare coefficients of `e^x`:

`B+2A-5B-5A+6B=0`
`2B-3A=0=>2B-3/2=0 =>B=+3/4`

So the PI should be:

`y=(1/2x +3/4)e^x`

The following link gives the full solution