(D4 + 2D3 – 3D2 – 4D + 4)y = 0
Its auxilary equation is
m4 + 2m3 – 3m2 – 4m + 4 = 0
⇒ (m – 1)(m3 + 3m2 – 4) = 0
⇒ (m – 1)2(m2 + 4m + 4) = 0
⇒ (m – 1)2(m + 2)2 = 0
⇒ m = 1, 1, –2, –2.
C.F. is y = (C1 + C2x)ex + (C3 + C4x)e-2x
Hence, solution of given differential equation is y = (C1 + C2x)ex + (C3 + c4x)e-2x.