Question

use the laplace transform to solve the initial value problem: y''+y=1, y(0)=2 and y'(0)=0

Answer 1

Answer:

y(s) = sin t + 2 cost

Step-by-step explanation:

given,

     y''  + y = 1,                       y(0) = 2  and  y'(0) = 0

applying Laplace transformation both side

s²y(s) - s y(o)- y'(0) + y(s) = L{1}

s²y(s) - 2 s + y(s)  = L{1}

y(s)(s² + 1 ) =  L{1}  + 2 s

$y(s)= L^{-1}(\dfrac{L(1)}{s^2 + 1})+2L^{-1}(\dfrac{s}{s^2+1})$

     y(s) = sin t + 2 cost

hence, the required solution of Laplace will be

     y(s) = sin t + 2 cost