To determine the solution of the initial value problemy″+4y=sin⁡t+uπ(t)sin⁡(t−π):y(0)=0,y′(0)=0.Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..

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Viktor Wiley · Accepted Answer

Applying the Laplace transform to both sides of the differential equationy″+4y=sin⁡t+uπ(t)sin⁡(t−π).⇒L{y″+4y}=L{sin⁡t+uπ(t)sin⁡(t−π)}⇒L{y″}+L{4y}=L{sin⁡t}+L{uπ(t0sin⁡(t−π)}By using L{fn(t)}=snF(s)−sn−1(0)−…−f(n−1)(0) ande−csG(s)=L{uc(t)g(t−c)},L{sin⁡t}=1s2+1⇒[s2Y(s)−sy(0)−y′(0)]+4Y(s)=1s2+1+e−πss2+1Using the initial conditions, y(0)=0 and y′(0)=0,⇒s2Y(s)+4Y(s)=1+e−πss2+1⇒Y(s)[s2+4]=1+e−πss2+1⇒Y(s)=1+e−πs(s2+1)(s2+4)By using the partial fraction 1+e−πs(s2+1)(s2+4)=13(1+e−πs)⌊1(s2+1)+1(s2+4)⌋⇒Y(s)=13(1+e−πs)⌊1(s2+1)+1(s2+4)⌋Applying the inverse Laplace transformation on both sides,

To determine the solution of the initial value problem {y}{''}+{4}{y}= sin{{t}}+{u}_{pi}{left({t}right)} sin{{left({t}-piright)}}: y(0) = 0, y'(0) = 0. Also, draw the graphs of the solution and of the forcing function and explain the relation between the solution and the forcing function..

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