Solve the following initial value problem by using Laplace transformy″+4y=4t+12,y(0)=4,y′(0)=0
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y″+4y=4t+12,y(0)=4,y′(0)=0Taking the Laplace transform on both sideL{y"}+4L{y}=4L{t}+12L{1}s2L{y}−sy(0)−y′(0)+4L{y}=41s2+121s(s2+4)L{y}−4s−0=4(3s+1s)(s2+4)L{y}=4(3s+1s+s)(s2+4)L{y}=4(3s+1+s2s)L{y}=4(s2+3s+1s(s2+4))=4(s2+4+3s−3s(s2+4))=4[1s+3s−3s(s2+4)]=4[1s+14(12s−12s(s2+4))]=4[1s+14(3s2−3s2+12s−12s(s2+4))]=41s+s(3s+12)−3(s2+4)s(s2+4)L{y}=4s+3s+12s2+4−3sL{y}=1s+3ss2+4+12s2+4y(t)=L−1{1s}+3L−1{ss2+4}+6L−1{2s2+4}y(t)=1+3cos(2t)+6sin(2t)
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