Question

6. tx" - 2x' + tx = 0, x(0) = 0.

Use the Laplace transform to solve initial value problems

Answer 1

Solution:

Notice first that taking t = 0 into the given equation tx" - 2.r' + tx = 0 we get 0 - 2x'(0) + 0 = 0. it follows that x'O) = 0. Applying the Laplace transform to the equation and using THEOREM 1 we get L-1{tx" - 2x' + tx} = 0 d (L{x"}) - 2sX(s) -X(s) = 0 ds ds d (s2 X (s)}) – 2sX(s) – X'(s) = 0 ds -2sX(s) - s'X'(s) - 2sX(s) - X'(s) = 0 -(sº+1)X'(s) - 45X(s) = 0 (s2 + 1)X'(s) + 4X(s) = 0 2+1 The last differential equation is separable and equivalent to X'(s) 4 X(S) The general solution of this equation is с X(S) = (s2 + 1)2

Taking the inverse Laplace transform we get z(t) = 2-1 e-{co2+1;} Csint - tcost) (s2 +

where C + 0 is an arbitrary constant.

Thank you