The general solution of differential equation y '' '= (α + β) y' + e ^ (α + β) x

The general solution of differential equation y '' '= (α + β) y' + e ^ (α + β) x

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Let a = α + β
(1) If a = α + β = 0
y'''=1
y''=x+A
y'=x^2/2+Ax+B
y=x^3/6+Ax^2/2+Bx+C
(2) Others
y'''=ay'+exp(a*x)
This is a linear ordinary differential equation
Finding homogeneous solution first
y'''=ay'
Let z = y '
z''=az
z''-az=0
The eigenvalue equation is
r^2-a=0
R = ± root a
Whether a is positive or negative, it's a big deal to get a complex number
Z = a'exp (radical a x) + b'exp (- radical a x) = y '
Y = aexp (radical a x) + bexp (- radical a x) + C
Another special solution
If a = α + β ≠ 0,1
Obviously, it can be assumed that y = h * exp (AX)
a^3h-a^2h=1
h=1/[a^2(a-1)]
That is y = aexp (radical a x) + bexp (- radical a x) + C + exp (AX) / [a ^ 2 (A-1)]