using laplace transform solve IVP y4yut2 y00y01SolutionGiven

using laplace transform solve IVP y\"+4y=u(t-2)

y(0)=0,y\'(0)=1

Solution

Given: y\'\'+4y=u(t-2) , y(0)=0,y\'(0)=1

           applying laplce transform on bothsides

        s²Y(s) - sy(0) - y\'(0) + 4Y(s) =e^-2s /s              [ from y(0)=0,y\'(0)=1]

         Y(s)(s²+4)-0-1 =e^-2s /s

          Y(s)(s²+4) =(e^-2s /s)+1

            Y(s) =[(e^-2s /s)+1]/(s²+4)

             Y(s) =[(e^-2s /s(s²+4))+(1/(s²+4))]

            Y(s)=e^-2s [(1/4)/(s)+(-1/4(s))/(s²+4))+(1/(s²+4))]

             Y(s)=[e^-2s (1/4)/(s)]+[e^-2s(-1/4(s))/(s²+4))]+[(1/(s²+4))]    

           apply inverse transform.

         y(t)=u(t-2)+(1/2)[u(t-2)cos[2(t-2)]]+(1/2)sin(2t).