Question-142910

Question Number 142910 by BHOOPENDRA last updated on 07/Jun/21

Answered by qaz last updated on 07/Jun/21

y (t) = t + e^{- 2 t} + \int_{0}^{t} y (τ) e^{2 (t - τ)} d τ

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

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

= \frac{2}{3} \cdot \frac{1}{s^{2}} + \frac{B}{s} + \frac{4}{5} \cdot \frac{1}{s + 2} + \frac{14}{45} \cdot \frac{1}{s - 3}

B = \lim_{s \to 0} \frac{d}{ds} [s^{2} L] = \lim_{s \to 0} \frac{d}{ds} [\frac{(s^{2} + s + 2) (s - 2)}{(s + 2) (s - 3)}]

= \lim_{s \to 0} \frac{[(2 s + 1) (s - 2) + (s^{2} + s + 2)] (s + 2) (s - 3) - (s^{2} + s + 2) (s - 2) (2 s - 1)}{{[(s + 2) (s - 3)]}^{2}}

= - \frac{1}{9}

\Rightarrow L = \frac{2}{3} \cdot \frac{1}{s^{2}} - \frac{1}{9} \cdot \frac{1}{s} + \frac{4}{5} \cdot \frac{1}{s + 2} + \frac{14}{45} \cdot \frac{1}{s - 3}

\Rightarrow y (t) = L^{- 1} = \frac{2}{3} t - \frac{1}{9} + \frac{4}{5} e^{- 2 t} + \frac{14}{45} e^{3 t}

Commented by BHOOPENDRA last updated on 07/Jun/21

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Answered by Olaf_Thorendsen last updated on 07/Jun/21

y (t) = t + e^{- 2 t} + e^{2 t} \int_{0}^{t} y (τ) e^{- 2 τ} d τ

y (0) = 0 + e^{0} + e^{0} \times 0 = 1

e^{- 2 t} y (t) = e^{- 2 t} t + e^{- 4 t} + \int_{0}^{t} y (τ) e^{- 2 τ} d τ

e^{- 2 t} (y^{'} (t) - 2 y (t)) = e^{- 2 t} (1 - 2 t)

- 4 e^{- 4 t} + y (t) e^{- 2 t}

y^{'} (t) - 3 y (t) = 1 - 2 t - 4 e^{- 2 t}

y (t) = \frac{2}{3} t - \frac{1}{9} + \frac{4}{5} e^{- 2 t} + {a e}^{3 t}

y (0) = - \frac{1}{9} + \frac{4}{5} + a = 1 \Rightarrow a = \frac{14}{45}

y (t) = \frac{2}{3} t - \frac{1}{9} + \frac{4}{5} e^{- 2 t} + \frac{14}{45} e^{3 t}

Commented by BHOOPENDRA last updated on 07/Jun/21

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