Define a solution the given initial-value problem.4″ + 4y′4y=δ(t − 4)y(0)=1y′(0)=2

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Define a solution the given initial-value problem.4″ + 4y′4y=δ(t − 4)y(0)=1y′(0)=2

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To solve, we use the properties of the Laplace transform and the Laplace tableThe Laplace transform satisfies the linearity propertiesL[f=g]=L[f] + L[g]L[cf]=cL[f]For all transfotmable functions f and g and constants c.L[y″ + 4y′ + 4y]=L[δ(t − 4)]L[y″] + 4L[y′] + 4Ly=L[δ(t − 4)]s2Y(s) − sy(0) − y′(0) + 4(sY(s) − y(0)) + 4Y(s)=e−4sY(s)(s2 + 4s + 4) − y′(0) − y(0)(s + 4)=e−4sSubstitute initial conditions,y(0)=1y′(0)=2Solve for Y(s).Y(s)(s2 + 4s + 4) − 2 − s − 4=e−4sY(s)=e−4s + s + 6(s + 2)2=e−4s(s + 2)2 + 1s + 2 + 4(s + 2)2Inverse transform:L−1[Y(s)]=L−1[e−4s(s + 2)2 + 1s + 2 + 4(s + 2)2]y(t)=L−1[e−4s 1(s + 2)2] + L−1[1s + 2] + 4L−1[1(s + 2)2]=L−1[e−4sL[te−2t]] + e−2t + 4te−2t=u4(t)(t − 4)e−2(t − 4) + e−2t(1 + 4t)

Define a solution the given initial-value problem. 4'' + 4y'4y=delta(t - 4) y(0) = 1 y'(0) = 2

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