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remolatg

Answered question

2021-02-08

Use Laplace Transform to solve the given equation:

y^{″} + 6 y^{'} + 5 y = t \cdot U (t - 2)

y (0) = 1

y^{'} (0) = 0

smallq9

Skilled2021-02-09Added 106 answers

Step 1
Let Laplace transform of y(t) be Y(s). Then Laplace transform of

y^{'} (t) and y " (t) is s Y (s) - y (0) or s Y (s) - 1 and s^{2} Y (s) - s y (0) - y^{'} (0) or s^{2} Y (s) - s

Also Laplace transform of

u (t - 2)

\frac{e^{- 2 s}}{s}

Therefore Laplace transform of

t u (t - 2)

\frac{- d}{d s} \frac{e^{- 2 s}}{s}

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

Step 2
Finally substituting the results in the equation given:

s^{2} Y (s) - s + 6 (s Y (s) - 1) + 5 Y (s) = \frac{(2 s + 1) e^{- 2 s}}{s^{2}}

(s^{2} + 6 s + 5) Y (s) - s - 6 = (2 s + 1) \frac{e^{- 2 s}}{s^{2}}

(s^{2} + 6 s + 5) Y (s) = \frac{(2 s + 1) e^{- 2 s}}{s^{2}} + s + 6

Y (s) = \frac{(2 s + 1) e^{- 2 s}}{s^{2} (s^{2} + 6 s + 5)} + \frac{s + 6}{s^{2} + 6 s + 5}

Y (s) = \frac{(2 s + 1) e^{- 2 s}}{s^{2} (s + 5) (s + 1)} + \frac{s + 6}{(s + 5) (s + 1)}

Y (s) = e^{- 2 s} (- \frac{\frac{32}{25}}{s} + \frac{\frac{7}{5}}{s^{2}} + \frac{\frac{5}{4}}{s + 1} + \frac{\frac{3}{100}}{s + 5}) + \frac{\frac{5}{4}}{s + 1} - \frac{\frac{1}{4}}{s + 5}

Step 3
Now Laplace inverse of:

\frac{1}{s} is u (t)

\frac{1}{s^{2}} is t u (t)

\frac{1}{s + 1} is e^{- t} u (t)

\frac{1}{s + 5} is e^{- 5 t} u (t)

Step 4
Thus using time shifting property and solving

Y (s) = e^{- 2 s} (- \frac{\frac{32}{25}}{s} + \frac{\frac{7}{5}}{s^{2}} + \frac{\frac{5}{4}}{s + 1} + \frac{\frac{3}{100}}{s + 5}) + \frac{\frac{5}{4}}{s + 1} - \frac{\frac{1}{4}}{s + 5}

y (t) = (- \frac{32}{25} u (t - 2)) + (\frac{7}{5} (t - 2) u (t - 2)) + (\frac{5}{4} e^{2 - t} u (t - 2)) + (\frac{3}{100} e^{10 - 5 t} u (t - 2)) + (\frac{5}{4} e^{- t} u (t)) - (\frac{1}{4} e^{- 5 t} u (t))

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