Can you please help me find the general solution for these problems:

\f\fWhen the auxiliary equation contains: Case 1: Distinct roots. The solution is y = Cjemi* + cemix + ... Cnemnx Case 2: Repeated roots. m1 = m2 = mn The solution is y = emx (c] + Czx + ... Coxn-1) Case 3: Imaginary roots m = a tib The solution is y = eax(c,cosbx + c2sinbx) Case 4: Repeated imaginary roots m1 = m2_a t ib The solution is y = eax[(C, + C2x + Cnx"-1)cosbx + (k, + K2x + knx"-1)sinbx]4. (D + 904 + 24D 2 + 16 ) 4 = 0 The Auxiliary Equation , mo + am4 + 24 m 2 + 16 = 0 let a = m2 a3 +aa2 + 240 + 16 = 0 By Synthetic Division, 1 9 24 16 1- 4 -4-20 -14 15 4 0 ( a + 4 ) ( 92 + 50 + 4 ) = 0 ( a+4 ) ( at1 ) ( at4 ) = 0 ( at 1 ) (at 4 ) 2 = 0 at1 = 0 , ( at4 ) ? = 0 since a = mz mz+1 = 0 , ( m 2+ 4 ) 2 = 0 m z = - 1 - we expect 4 roots from this equation. m = IL 9=0 MI =L m3 = MA = 21 7 9= 0 6 = 1 M 2 = - L m5 = M6 = - 2L b = 2 case 3 case 4 solution : y = e ( c , cos x + c 2 sinx ) + 20 [ ( c3 + CA ) cos 2x + ( C5 + Cox)sin2x