4 5(10pt). Use Laplace transform to solve the equation: (3) + x = 0, x(0) = 0, r'(0) = 0.1"(0) = 1. - 6(10pt). Find the inverse Laplace transform of F(s) using Theorem 1 on page 298. and that is, (cos) (sin7) sin-sin(2x-1)) dr sin+cos(2-1)] (cos 1) (sin) sin. And we recall from Example 5 of Section 4.2 that the Laplace transform of is indeed s/(s2+1)². Theorem 1 is proved at the end of this section. THEOREM 1 The Convolution Property and ig()) are bounded by Me" as t+oo. Then the Laplace transform of the Suppose that f(t) and g(t) are piecewise continuous for 10 and that f convolution f(r)g(t) exists for s> c; moreover, L[f(t) = g(1)) = L{f(t)) L{g(t)} (4) L(F(s) G(s)) = f(t) * g(t). (5) Thus we can find the inverse transform of the product F(s) G(s), provided that we can evaluate the integral L(F(s). G(s)) = f(t)g(tt)dt. (5) Example 2 illustrates the fact that convolution often provides a convenient alternative to the use of partial fractions for finding inverse transforms. With f(t)=sin 21 and g(t) = e', convolution yields 2 e = ['e- = ['e- -D² + 4)] = (sin 21)+ sin 2r de 1)(s²+4) Example 2 L-1