using laplace transformation

solve d^2y/dt + dy/dt = t^2 + 2t

Given that y(0) = 4, y'(o)=2

Dear KHITISH

d²y/dt² + dy/dt = t^2 + 2t

take laplace

[s² L(y) -sy(0) -y'(0) ] + [sL(y)-y(o)] =2!/s³    + 2/s²

[s² L(y) -4s -2 ] + [sL(y)-4] =2!/s³    + 2/s²

 [s² +s]L(y) =2!/s³    + 2/s²  +4s +6

         L(y) =2/s⁴(s+1)   + 2/s³(s+1)   + 4/(s+1)      +6/s(s+1)

               =2/s⁴  +6/s -2/(s+1)

now take inverse

   y = 2t³/3!  + 6   -2e^-t

        =t³/3 + 6   -2e^-t

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