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anonymous

  • one year ago

how to solve this ODE: I will post it

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  1. anonymous
    • one year ago
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    \[\frac{ d ^{2} x }{ dt } = 32 - \frac{ v^2 }{ 8 }\]

  2. anonymous
    • one year ago
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    I tried to solve it for v then for x, but it was very messy.

  3. anonymous
    • one year ago
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    @e.mccormick

  4. anonymous
    • one year ago
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    @phi

  5. anonymous
    • one year ago
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    \[\frac{ d }{ dt }\left( \frac{ dx }{ dt } \right)=32-\frac{ v^2 }{ 8 }\] \[\frac{ dv }{ dt }=\frac{ 256-v^2 }{ 8 }\] \[\frac{ dv }{ 256-v^2 }=8 dt\] \[\frac{ dv }{ \left( 16-v \right)\left( 16+v \right) }=8 dt\] \[\left\{ \frac{ 1 }{\left( 16-v \right)\left( 16+16 \right) } +\frac{ 1 }{ \left( 16+16 \right)\left( v+16 \right) }\right\} dv=8dt\] can you solve further?

  6. anonymous
    • one year ago
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    of course you don't need to make partial fraction as you can sub.with sin theta and you will get ln(sec+tan) +k and then integrate which that!! my text didn't discuss DE at all just basic integral, but he gave that equation and said after 1 minute it will hit the ground, so find height??(ans 555) I am puzzled about what is his simple way to deal with that?

  7. anonymous
    • one year ago
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    when t=1 sec. when it touches the ground v=0 find the value of constant then again integrate to find x when t=1 x (height)=0 find value of constant.

  8. anonymous
    • one year ago
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    sorry 1 min. just take t=1

  9. IrishBoy123
    • one year ago
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    @Catch.me why don't you just scan or link the actual question? the DE is not the problem, it is quite straightforward. sight of the question and the IV's would be interesting. @phi

  10. IrishBoy123
    • one year ago
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    and, as a complement to @surjithayer 's excellent input: \[\ddot x = \frac{dv}{dt} = \ v \frac{dv}{dx}\] :p

  11. anonymous
    • one year ago
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    V final can't be zero as it is a different scenario. I solved it, but 60 in expm. worries me. any way thanks guys

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