To determine the solution of the initial value problemy″+4y=δ(t−π)−δ(t−2π),y(0)=0,y′(0)=0 and draw the graph of the solution.

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Nola Robson · Accepted Answer

Decision:Let the differential equation y″+4y=δ(t−π)−δ(t−2π).Applying the Laplace transform to the differential equationL{y″+4y}=L{δ(t−π)−δ(t−2π)}L{y″}+4L{y}=L{δ(t−π)}−L{δ(t−2π)}By using L{fn(t)}=snF(s)−sn−1f(0)−…−fn−1(0) andL{δ(t−t0)}=est0,⇒[s2Y(s)−sy(0)−y′(0)]+4Y(s)=e−πs−e−2πsUsing the initial conditions y(0)=0 and y′(0)=0,s2Y(s)+4Y(s)=e−πs−e−2πs⇒Y(s)[s2+4]=e−πs−e−2πs⇒Y(s)=e−πs−e−2πss2+4Now, applying the inverse Laplace transformation on both sides,⇒L−1{Y(s)}=L−1{e−πss2+4−e−2πss2+4}⇒L−1{Y(s)}=12L−1{2e−πss2+22}−12L−1{2e−2πss2+22}By using L−1{ecsG(s)}=uc(t)g(t−c) and properties of inverse Laplace,

To determine the solution of the initial value problem {y}{''}+{4}{y}=delta{left({t}-piright)}-delta{left({t}-{2}piright)},{y}{left({0}right)}={0},{y}'{left({0}right)}={0} and draw the graph of the solution.

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