Solve the differential equations by Laplace transform

Question

a) 𝑦′′(𝑡) + 𝑦′ (𝑡) + 3𝑦(𝑡) = 0, 𝑦(0) = 1, 𝑦′ (0) = 2

b) 𝑦′′(𝑡) + 𝑦′ (𝑡) = 𝑠𝑖𝑛𝑡, 𝑦(0) = 1, 𝑦′ (0) = 2

Answer 1

a) s²F(s) - s - 2 + sF(s) - 1 + 3F(s) = 0, F(s)(s² + s + 3) = s + 3,·F(s) = (s + 3)/(s² + s + 3) =(s + 1/2)/[(s + 1/2)²+ 11/4] + 5/2/[(s+ 1/2)² + 11/4}

Now inverse Laplace transform give us: y = e^-1/2t[cos(11^1/2t/2) + 5/11^1/2sin(11^1/2t/2)]

b) s²F(s) - s - 2 + sF(s) - 1 + 3F(s) = 1/(s² + 1); F(s)(s² + s+ 3) = 1/(s² + 1) + s + 3

So F(s) = (s+3)/(s²+ s+ 3) + 1/[(s² + s + 3)(s² + 1)]

We ned decomposition only for 1/[(s² + s + 3)(s² + 1)] = (as + b)/(s² + s + 3) + (cs + d)/(s² + 1);

a = 0, c = 0, b = -1/2, d = 1/2.

We get F(s) = (s + 3)/(s² + s + 3) - 1/2/[(s + 1/2)² + 11/4] + 1/2/(s² + 1)

So, y = e^-1/2t[cos(11^1/2t/2) + 5/11^1/2sin(11^1/2t/2)] + 1/2sint - 1/11^1/2e^-1/2tsin(11^1/2t)