Question

solve the following using la place
d^2x/dt^2 +2dx/dt +5x = 1
given t =0 , x =0 and dx/dt =0

Answer 1

Have you done any work with this one yet?

Answer 2

yeah i have but the problem i dont know how to apply the 3rd condition dx/dt =0

Answer 3

dx/dt (0) = 0
becomes useful when you use properties of Laplace transform to convert the transform of the second derivative of x.

Answer 4

\( \displaystyle \mathcal{L} \left[ \frac{d^2 x}{dt^2 } \right] = s \mathcal{L} \left[ \frac{dx}{dt} \right] - \frac{dx}{dt} (0) \)

Answer 5

This just being a consequence of the case with a first derivative.

Answer 6

so can u show me how in my equation

Answer 7

Well, you take the Laplace transform of both sides to start.
\( \displaystyle \mathcal{L} \left[ \frac{d^2 x}{dt^2} + 2 \frac{dx}{dt} + 5x \right] = \mathcal{L} \left[ 1 \right] \)
Linearity of the Laplace transform operator lets you take the Laplace transform term-by-term and factor out constants freely.
\( \displaystyle \mathcal{L} \left[ \frac{d^2 x}{dt ^2 } \right] + 2 \mathcal{L} \left[ \frac{dx}{dt} \right] + 5 \mathcal{L} \left[ x \right] = \mathcal{L} \left[1 \right] \)
That part is clear, correct?

Answer 8

yeah

Answer 9

This property is then useful to us, concerning the derivatives of x:
\( \displaystyle \mathcal{L} \left[ \frac{dx}{dt} \right] = s \; \mathcal{L} \left[ x \right] - x(0) \)

Answer 10

We can use it for the second derivative as well, simply calling the second derivative "the derivative of dx/dt".

\( \displaystyle \mathcal{L} \left[ \frac{d^2 x }{dt^2 } \right] = s \mathcal{L} \left[ \frac{dx}{dt} \right] - \frac{dx}{dt} (0) \)
and then using the original property for the laplace transform of dx/dt.
\( \displaystyle = s \left( s \mathcal{L} \left[ x \right] - x(0) \right) - x'(0) \)
\( \displaystyle = s^2 \mathcal{L} \left[ x \right] - s x(0) - x'(0) \)
Does all of that seem clear then?

Answer 11

yeah it is clear

Answer 12

Alright, so we can directly substitute those two cases into what we have above:
\( \displaystyle \mathcal{L} \left[ \frac{d^2 x}{dt ^2} \right] + 2 \mathcal{L} \left[ \frac{dx}{dt} \right] + 5 \mathcal{L} \left[ x \right] = \mathcal{L}\left[ 1 \right] \)
\( \displaystyle (s^2 \mathcal{L} \left[x \right] - s x(0) - x'(0) )+ 2 (s \mathcal{L} \left[ x \right] - x(0)) + 5 \mathcal{L} \left[ x \right] = \frac{1}{s} \)
1/s is the Laplace transform of 1.
If this all makes sense, you should be able to substitute all the conditions and get the answer. :)

Answer 13

all this it is make sense for me but how can apply the dx/dt = 0

Answer 14

x'(0) = 0 is the same as dx/dt = 0 at t=0. Just a bit of ease of notation.

Answer 15

When they say,
"given t =0 , x =0 and dx/dt =0"
x = 0 at t = 0 x(0) = 0
and dx/dt = 0 at t = 0. x ' (0) = 0.

Answer 16

okayyyyyyyyyyyyyyyyyyyyy now i got it tanxxxxxxx so much sir

Answer 17

You're welcome! :)