Question

Solve the intial value problem.
d^2r/dt^2=3/t^4; dr/dt t=1 =1, r(2)=2
Please Help!

Answer 1

\[\frac{d^2r}{dt^2}=\frac{3}{t^4} \\ \frac{d}{dt}(\frac{dr}{dt})=\frac{3}{t^4} \\ hence \ \ by \ \ integration \\ \frac{dr}{dt}=-\frac{1}{t^3}+c\]

Answer 2

am i clear?

Answer 3

yes correct

Answer 4

now u have
dr/dt t=1 =1
put this bondary condition in the equation
what will u get?

Answer 5

i put the boundary into dr/dt correct?

Answer 6

C=1?

Answer 7

thats right

Answer 8

now u have
\[\frac{dr}{dt}=-\frac{1}{t^3}+1\]
integrate again to get the final answer

Answer 9

this is where i have a problem where i have to intergrate again

Answer 10

1/3t^2+1t+C?

Answer 11

1/2t^2+1t+C

Answer 12

oh ok i eas really wrong

Answer 13

i meant was not eas

Answer 14

u have to integrate until u have such terms that include derivatives of your unknown function

Answer 15

oh ok

Answer 16

what is final answer?

Answer 17

so is the answer 1/2t^2+1t-1/8 ?

Answer 18

yes thats completely right

Answer 19

ok thanks so much for your help =)

Answer 20

welcome