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

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Answer:

y(s) = sin t + 2 cost

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)((L(1))/(s^2 + 1))+2L^(-1)((s)/(s^2+1))

y(s) = sin t + 2 cost

hence, the required solution of Laplace will be

y(s) = sin t + 2 cost

User Louis Caron
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