Use-Laplace-transform-to-solve-the-differential-equation-d-2-v-t-dt-2-6-dv-t-dt-8v-t-2u-t-when-v-0-1-and-v-0-2-

Question Number 184936 by MikeH last updated on 14/Jan/23

Use Laplace transform to solve the differential

equation

\frac{d^{2} v (t)}{{d t}^{2}} + 6 \frac{d v (t)}{d t} + 8 v (t) = 2 u (t)

when v (0) = 1 and \overset{∙}{v} (0) = - 2

Answered by hmr last updated on 15/Jan/23

L {v^{″}} + 6 L {v^{'}} + 8 L {v} = 2 L {u}

\to

(s^{2} L {v} - s v (0) - v^{'} (0))

+ 6 (s L {v} - v (0))

+ 8 L {v} = 2 L {u}

\to

(s^{2} + 6 s + 8) L {v} - s + 2 - 6 = 2 L {u}

\to

(s^{2} + 6 s + 8) L {v} = 2 L {u} + s + 4

\to

L {v} = \frac{2 L {u} + s + 4}{s^{2} + 6 s + 8}

\to

L {v} = \frac{2 L {u} + (s + 4)}{(s + 4) (s + 2)}

\to

L {v} = \frac{2 L {u}}{(s + 4) (s + 2)} + \frac{1}{s + 2}

\to

L {v} = \frac{2 L {u}}{(s + 4) (s + 2)} + L {e^{- 2 t}}

\to

L {v} = \frac{4 L {u}}{s + 4} - \frac{2 L {u}}{s + 2} + L {e^{- 2 t}}

\to

L {v} = 4 L {e^{- 4 t}} L {u} - 2 L {e^{- 2 t}} L {u} + L {e^{- 2 t}}

S o r r y {t h a t}^{'} s a l l I k n o w .

Commented by MikeH last updated on 15/Jan/23

thanks a lot

Commented by hmr last updated on 15/Jan/23

{Y o u}^{'} r e W e l c o m e

Commented by hmr last updated on 15/Jan/23

I e d i t e d t h e l a s t l i n e, I t w a s w r o n g!

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