Chapter 5 Integration Ex 5.3 Solutions

Let I = Put, Numerator = A(Denominator) + B[(Denominator)] ∴ 3e2t + 5 = A(4e2t – 5) + B[(4e2t – 5)] ∴ 3e2t + 5 = A(4e2t – 5) + B[4e2t × 2 – 0] ∴ 3e2t + 5 = (4A + 8B) e2t – 5A Equating the coefficient of e2t and constant on both sides, we get 4A + 8B = 3 and -5A = 5 ∴ A = -1 ∴ from (1), 4(-1) + 8B = 3 ∴ 8B = 7 ∴ B =

Let I = Put, Numerator = A (Denominator) + B[(Denominator)] ∴ 20 – 12ex = A(3ex – 4) + B[(3ex – 4)] ∴ 20 – 12ex = A(3ex – 4) + B(3ex – 0) ∴ 20 – 12ex = (3A + 3B)ex – 4A Equating the coefficient of ex and constant on both sides, we get 3A + 3B = -12 ……(1) and -4A = 20 ∴ A = -5 from (1), 3(-5) + 3B = -12 ∴ 3B = 3 ∴ B = 1 ∴ 20 – 12ex = -5(3ex – 4) + (3ex)

Let I = Put, Numerator = A (Denominator) + B[(Denominator)] ∴ 3ex + 4 = A(2ex – 8) + B[(2ex – 8)] ∴ 3ex + 4 = A(2ex – 8) + B(2ex – 0) ∴ 3ex + 4 = (2A + 2B)ex – 8A Equating the coefficient of ex and constant on both sides, we get 2A + 2B = 3 ……..(1) and -8A = 4 ∴ A = ∴ from (1), 2() + 2B = 3 ∴ 2B = 4 ∴ B = 2