Find the solution of the initial value problem given below by Laplace transform y'-y=t e^t sin t y(0)=0

generals336 2021-03-06 Answered

Find the solution of the initial value problem given below by Laplace transform

y^{'} - y = t e^{t} \sin t

y (0) = 0

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Expert Answer

stuth1

Answered 2021-03-07 Author has 97 answers

Step 1
The answer is given by using properties of Laplace transform.
Step 2

y^{'} - y = t e^{t} \sin (t)

y (0) = 0

Apply the laplace transform

L {y^{'}} - L {y} = L {t e^{t} \sin (t)}

\Rightarrow (s Y (s) - y (0)) - Y (s) = \frac{2 (s - 1)}{(s^{2} - 2 s + 2)^{2}}

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

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

\Rightarrow Y (s) = \frac{2}{(s^{2} - 2 s + 2)^{2}}

Now apply inverse laplace transform

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

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

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

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

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