# 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#?

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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. :)

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# .

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. :)

I was letting

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

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

#A_0 = 1/2#

Constant terms:

##### 2 Answers

The solution

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

as can be easily verified, so the complete solution is

I'll use

The problem is you are trying the wrong solution you have

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

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

Compare coefficients of

#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#