Question

how to solve this ODE:
I will post it

Answer 1

\[\frac{ d ^{2} x }{ dt } = 32 - \frac{ v^2 }{ 8 }\]

Answer 2

I tried to solve it for v then for x, but it was very messy.

Answer 3

@e.mccormick

Answer 4

@phi

Answer 5

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

Answer 6

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?

Answer 7

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.

Answer 8

sorry 1 min.
just take t=1

Answer 9

@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

Answer 10

and, as a complement to @surjithayer 's excellent input:

\[\ddot x = \frac{dv}{dt} = \ v \frac{dv}{dx}\]



:p

Answer 11

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