Second order differential equation

y'' -4y' +3y = 4e^3x, y(0) = 1 y'(0) = 1

y'' - 4y' + 3y = 4e^3x, y(0) = 1 y'(0) = 1

complimentary solution y₁:
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r² - 4r + 3 = 0
r² - 3r - r + 3 = 0
r(r - 3) - 1(r - 3) = 0
r = 1,3
y₁ = C₁e^x + C₂e^(3x)

particular solution y₂;
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according to the right-hand side of the differential equation y'' -4y' +3y = 4e^3x this would suggest that y₂ be in the form , y₂ = Ae^(3x) but homogeneous solution already e^(3x). multiply it by x to make this solution unique, so y₂ becomes y₂ = Axe^(3x)

y₂ = Axe^(3x)
y₂' = 3Axe^(3x) + Ae^(3x)
y₂'' = 9Axe^(3x) + 6Ae^(3x)

y'' - 4y' + 3y = 4e^3x
9Axe^(3x) + 6Ae^(3x) - 4( 3Axe^(3x) + Ae^(3x) ) + 3( Axe^(3x) ) = 4e^(3x)
2Ae^(3x)= 4e^(3x)
A = 2

y₂ = 2xe^(3x)

general solution
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y = y₁ + y₂
y = C₁e^x + C₂e^(3x) + 2xe^(3x)

initial conditions:
y = C₁e^x + C₂e^(3x) + 2xe^(3x)
y(0) = C₁ + C₂ = 1
C₁ + C₂ = 1

y' = C₁e^x + 3C₂e^(3x) + 6xe^(3x) + 2e^(3x)
y(0)' = C₁ + 3C₂ + 2 = 1
C₁ + 3C₂ = -1

C₁ = 2
C₂ = -1
y = 2e^x - e^(3x) + 2xe^(3x)

you can simplify this by factoring out 'common factors'
y = 2e^x - e^(3x) + 2xe^(3x)
y = e^x [ 2 - e^(2x) + 2xe^(2x) ]
y = e^x [ 2 - e^(2x) [1 + 2x] ]
y = [ 2 - (1 + 2x)e^(2x) ]e^x
y = [ 2 + (2x - 1)e^(2x) ]e^x

you certainly can do this...you know y_c = c1 e^x + c2 e^(3x) ..so y_p = A x e^(3x) for some A