Use Laplace transform to solve the folowing initial value problem y"+2y'+2y=0 y(0)=2 y'(0)=-1

Question

Use Laplace transform to solve the folowing initial value problem

y " + 2 y^{'} + 2 y = 0

y (0) = 2

y^{'} (0) = - 1

Answer 1

Step 1Given

y " + 2 y^{'} + 2 y = 0

y (0) = 2

y^{'} (0) = - 1

Taking Laplace transform on both sides

L {y "} + 2 L {y^{'}} + 2 L {y} = 0

\Rightarrow s^{2} y (s) - s y (0) - y^{'} (0) + 2 [s y (s) - y (0)] + 2 y (s) = 0

given

y (0) = 2

y^{'} (0) = - 1

\Rightarrow s^{2} y (s) - 2 s + 1 + 2 [s y (s) - 2] + 2 y (s) = 0

\Rightarrow [s^{2} + 2 s + 2] y (s) - 2 s - 3 = 0

\Rightarrow y (s) = \frac{2 s + 3}{s^{2} + 2 s + 2}

Step 2

L^{- 1} y (s) = L^{- 1} {\frac{2 s + 3}{s^{2} + 2 s + 2}}

= L^{- 1} {2 \frac{s + 1}{(s + 1)^{2} + 1} + 1 \frac{1}{(s + 1)^{2} + 1}}

= 2 L^{- 1} {\frac{s + 1}{(s + 1)^{2} + 1}} + L^{- 1} {\frac{1}{(s + 1)^{2} + 1}}

\Rightarrow y (t) = 2 e^{- t} \cos t + e^{- t} \sin t

\Rightarrow y (t) = 2 e^{- t} \cos t + e^{- t} \sin t

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