Question

Can someone help me with this initial value problem? d^2y over dt^2 = 2/t^3. Given is: dr/dt = t=1 and that whole thing = 1. Another Given is: r(1) = 1. I have no idea how to go about this problem after finding the antiderivative for the d^2r over dx^2.

Answer 1

so you never meant to say anything about y right?

Answer 2

Sorry meant r not y.

Answer 3

\[r''=2t^{-3}\]

Answer 4

So you said you integrated both sides:
\[r'=2 \cdot \frac{t^{-3+1}}{-3+1}+C_1=2 \cdot \frac{t^{-2}}{-2}+C_1=-1 t^{-2}+C_1\]
=>
\[r'=\frac{-1}{t^2}+C_1\]

Answer 5

So we are given r'(1)=1?

Answer 6

Just r(1)=1

Answer 7

dr/dt = t=1?
r'=t=1
I look at that is when t=1 ,r '=1

Answer 8

\[1=\frac{-1}{1^2}+C_1 \]
Solve for C_1

Answer 9

where does the r' come from?

Answer 10

Oh! so another way to look at d^2r/dt^2 is r''?

Answer 11

yep yep

Answer 12

Okay! Also just to clarify, when substituting t and r, just plug in one in newly acquired derivative where r and t are respectively? If r(1)=1 and t(1)=2, then plug in respectively and then ... what would happen?

Answer 13

well if you are given r'(a)=b
then you use the function for r'
If t=a then r'=b
------------------------
and if you are given r(c)=d
then use the function r
If t=c, then r=d

You will use both of this conditions to find your constants

Answer 14

So, since r(1)=1, then t(1)=2, then r of t = 1? Somehow I confused myself...

Answer 15

t(1)=2 makes no sense

Answer 16

I'm sorry to be so blunt

Answer 17

No, it makes no sense to me either.

Answer 18

You are given r'(1)=1 and r(1)=1