differential & linear algebra question
!!! 3AM stress!

Question

differential & linear algebra question
!!! 3AM stress!

adunb8 · Answer

$$2t(d^2x/dt^2) = (dx/dt)^2 $$

adunb8 · Answer

have to turn it into 1st order diff equation.

anonymous · Answer

ISn't it quite evident - you have derivative-of-a-derivative. Sooo if u want 1-st order
choose as your variable$$u = \frac{ dx }{ dt }$$

anonymous · Answer

??

adunb8 · Answer

i see i just got that part and got up to $$\int\limits_{}^{} 1/(v^2-1) dv = \int\limits_{}^{}(1/2t) dt$$

anonymous · Answer

seems Too final - can you show the intermediate stages ?

anonymous · Answer

It is unclear to me where did you get the "1" in the left hand integral

anonymous · Answer

@adunb8  ?

adunb8 · Answer

okay hold

adunb8 · Answer

i forgot that there was -1 at the right side

anonymous · Answer

So my guidance was useful. what do you say ?

adunb8 · Answer

yea but the problem is im not getting my teacher's answer..
its suppose to be $$x=-t-2/cln[ct-1]+k$$

anonymous · Answer

You did half the soln. you found x' - now integrate that once more.
And - if you are normal human, put a medal where medal is due.

adunb8 · Answer

im stuck here ... not getting the answer...

adunb8 · Answer

i did integrated and got ln |v^2-1| = ln|2t| then cancelled each out using e^
and left with v^2-1 = 2t + C

anonymous · Answer

on the left hand side you have to do partial fraction like this$$\frac{ 1 }{ \left( u-1 \right)\left( u+1 \right) }=\frac{ A }{ \left( u-1 \right) }+\frac{ B }{ \left( u+1 \right) }$$then solving for A and B and continue do integral you'll get two terms of ln

anonymous · Answer

It is all small technical steps - do them with precision.

adunb8 · Answer

oh i see partial fraction ??

anonymous · Answer

the final solution in the term of u, you'll get this$$\frac{ u+1 }{ u-1}=ct$$then de integral angain, since you've defined u = dx/dt

adunb8 · Answer

dx/dt + 1 / dx/dt -1

anonymous · Answer

$$\int\limits_{}^{}dx=-\int\limits_{}^{}\frac{ \left( 1+ct \right) }{ \left( 1-ct \right) }dt$$

anonymous · Answer

It's OK for you now ??

adunb8 · Answer

yeah thank for the clear instruction =)

anonymous · Answer

no problem sir :)

anonymous · Answer

I like you @Pinguine

adunb8 · Answer

me to i love you both =)

anonymous · Answer

><"

anonymous · Answer

much ♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥♥ love 
♥♥♥♥♥♥♥♥♥everybody♥♥♥♥♥♥♥♥