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Hence the solution is y = e3x ( A Cos2x + B Sin2x) 2. Solve (D2 – 2D + 1) y = ex. A.E is m2 – 2m + 1 =0 (m – 1)2 = 0 => m = 1,1 . C.F => (A + Bx) ex. P.I = Solution is y = C.F + P.I = (A + Bx) ex + 3. Find the particular integral of ( D2 + 5D + 6) y = 11 e5x P.I = Put D = 5 hence P.I = 4 Solve (D – 1) 2 y = sinh2x. A.E is (m – 1)2 = 0 ... ................
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