The solution of initial value problem y″ + 2y′ + 2y=δ(t − π), y(0)=0, y′(0)=1 and draw the graph of the solution.

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The solution of initial value problem y″ + 2y′ + 2y=δ(t − π), y(0)=0, y′(0)=1 and draw the graph of the solution.

Brighton · Accepted Answer

Let the differential equation y″ + 2y′ + 2y=δ(t − π)Applying the Laplace transform to the differential equation,Ly″ + 2y′ + 2y=Lδ(t − π)Ly″+2Ly′+2Ly=Lδ(t − π)By using Lfn(t)=snF(s) − sn − 1f(0) − ⋯ − fn − 1(0) andLδ(t − t0)=e−st0⇒ [s2Y(s) − sy(0) − y′(0)] + 2[sY(s) − y(0)] + 2Y(s)=e−π sUsing the initial conditios y(0)=0 and y′(0)=1,[s2Y(s) − 1] + 2sY(s)=eπ s⇒ Y(s)[s2 + 2s + 2] − 1=eπ s⇒ Y(s)=1 + eπ ss2 + 2s + 2⇒ Y(s)=1(s + 1)2 + 1 + e−π s(s + 1) + 1Now, applying the inverse Laplace transformation on both sides,⇒ L−1Y(s)=L−1{1(s + 1)2 + 1 + e−π s(s + 1)2 + 1}⇒ L−1Y(s)=L−1{1(s + 1)2 + 1} + L−1{e−π s(s + 1)2 + 1}By using L−1e−csG(s)=uc(t)g(t − c) and properties of inverse Laplace,⇒ y(t)=e−1sin⁡ t + uπ(t)e−(t − π)sin⁡(t − π)Therefore, the solution of the initial value problemy″ + 2y′ + 2y=δ(t − π), y(0)=0, y′(0)=1 isy(t)=e−tsin⁡ t + uπ(t)e−(t − n)sin⁡(t − π).By using

The solution of initial value problem y'' + 2y' + 2y=delta(t - pi), y(0)=0, y'(0)=1 and draw the graph of the solution.

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