Tricky Advanced Calc Question

bhaktir

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Aug 10, 2011
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The functions y1(t) and y2(t) are both solutions of the same autonomous differential equation dy/dt = 3sin(y/2) but satisfy different initial conditions: y1(0) = 1 and y2(0) = −1. Either by solving the differential equation or, better, by thinking about its geometry (slope field), calculate:

lim [(y1(t) − y2(t)] t→∞

(a) 0 (b) 2π (c) 4π (d) 6π (e) 8π (f) ∞

Answer: C

* π = pie symbol = 3.1459...
-----

Please help. My answer is NOWHERE near 4π!

Thank you!
 
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The functions y1(t) and y2(t) are both solutions of the same autonomous differential equation dy/dt = 3sin(y/2) but satisfy different initial conditions: y1(0) = 1 and y2(0) = −1. Either by solving the differential equation or, better, by thinking about its geometry (slope field), calculate:

lim [(y1(t) − y2(t)] t→∞

(a) 0 (b) 2π (c) 4π (d) 6π (e) 8π (f) ∞

Answer: C

* π = pie symbol = 3.1459...
-----

Please help. My answer is NOWHERE near 4π!

Thank you!

Start by solving the DE then think through geometric interpretation.

Please share your work with us, so that we may know where to begin to help you.
 
\(\displaystyle \int csc(x) dx \ = \ -ln[|csc(x) + cot(x)|] \ + \ C\)
 
So, this is how I tried to solve it. 1/(3sin(y/2)) on one side and nothing on the other. Turns out to be,

=>∫ 3csc(y/2) dy = t + c

=> -3ln [csc(y/2) + cot (y/2)] = t + c

=> csc(y/2) + cot (y/2) = - (Ae^t)/3

=> Input y and t values...

...And there I get stuck. Calculators aren't allowed in my class, but even with one I doubt I could attempt this, as the solution is nowhere near the answer I'm getting. And I can't seem to construct the solution plane for this either. I learn best by analyzing steps in example questions, so if someone could please explain each towards the solution, I'd be extremely grateful!

Please! My mid-term is in less than 5 hours. :(
 
If you solve the DE you get using y(0)=1:

\(\displaystyle y=4tan^{-1}\left(e^{\frac{3t}{2}+\frac{c}{2}}\right)+k\pi\)

Taking the limit as \(\displaystyle t\to \infty\) gives \(\displaystyle \lim y_{1}=2(2C+1)\pi\)

\(\displaystyle \lim y_{2}\) turns out to be \(\displaystyle y_{2}=2(2C-1)\pi\)

Subtracting them gives \(\displaystyle y_{1}=y_{2}=4\pi\)
 
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