Velocity of a particle large and small times

[rock.freak667]

Homework Statement

The velocity,v, of a particle moving on the x-axis varies with the time,t,according to the differential equation \frac{dv}{dt}+2v=sint. Find v in terms of t given that v=0 when t=0. Deduce that when t is large;the velocity of the particle varies between -0.45 and 0.45. Show that is t is very small,then v\approx\frac{1}{2}t^2

Homework Equations

\frac{d}{dx}(\frac{e^{2t}}{5}(2sint-cost))=e^{2t}sint

The Attempt at a Solution

\frac{dv}{dt}+2v=sint

X e^{2t}

e^{2t}\frac{dv}{dt}+2e^{2t}v=e^{2t}sint

\int...dt

ve^{2t}=\frac{e^{2t}}{5}(2sint-cost)+C

using the initial values they gave me C=\frac{1}{5}

And so
v=\frac{1}{5}(2sint-cost)+\frac{e^{-2t}}{5}

As t\rightarrow\infty(large),\frac{e^{-2t}}{5}\rightarrow0

but \frac{1}{5}(2sint-cost) does not approach any limit as both cost and sint oscillate between 1 and -1.

 

[real10]

Homework Statement

The velocity,v, of a particle moving on the x-axis varies with the time,t,according to the differential equation \frac{dv}{dt}+2v=sint. Find v in terms of t given that v=0 when t=0. Deduce that when t is large;the velocity of the particle varies between -0.45 and 0.45. Show that is t is very small,then v\approx\frac{1}{2}t^2

Homework Equations

\frac{d}{dx}(\frac{e^{2t}}{5}(2sint-cost))=e^{2t}sint

The Attempt at a Solution

\frac{dv}{dt}+2v=sint

X e^{2t}

e^{2t}\frac{dv}{dt}+2e^{2t}v=e^{2t}sint

\int...dt

ve^{2t}=\frac{e^{2t}}{5}(2sint-cost)+C

using the initial values they gave me C=\frac{1}{5}

And so
v=\frac{1}{5}(2sint-cost)+\frac{e^{-2t}}{5}

As t\rightarrow\infty(large),\frac{e^{-2t}}{5}\rightarrow0

but \frac{1}{5}(2sint-cost) does not approach any limit as both cost and sint oscillate between 1 and -1.

no not between -1 and 1 that is for individual terms.You have to combine them into 1 cosine term and the new amplitude is ur answer.
use the below formula:
A sint + B cost = (A^2+B^2 ) ^(1/2) * Cos(t-arctan(b/A))

 

[rock.freak667]

but wouldn't Cos(t-arctan(b/A)) not approach a limit? Or would t tending to infinity mean that t would eventually be equal toarctan(B/A) and so it would be the cos0=1?

 

[bob1182006]

the final limit will include a function that oscillates between some numbers.
since that function will oscillate it means that the velocity will vary.
Between [-0.45,0.45] in this case.
You will have to plug in the numbers that lead to the maximum value (0.45) and the minimum (-0.45). and that will be it for t->infinity.

 

[rock.freak667]

Oh...thank you...shall try the next part now

 

[real10]

Oh...thank you...shall try the next part now

yes it is just the amplitude as I said [( 2/5)^2 + (1/5)^2 ] ^1/2 = +-0.45