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sharathe
 3 years ago
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.
sharathe
 3 years ago
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.

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freckles
 3 years ago
Best ResponseYou've already chosen the best response.1so you never meant to say anything about y right?

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1So you said you integrated both sides: \[r'=2 \cdot \frac{t^{3+1}}{3+1}+C_1=2 \cdot \frac{t^{2}}{2}+C_1=1 t^{2}+C_1\] => \[r'=\frac{1}{t^2}+C_1\]

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1So we are given r'(1)=1?

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1dr/dt = t=1? r'=t=1 I look at that is when t=1 ,r '=1

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1\[1=\frac{1}{1^2}+C_1 \] Solve for C_1

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0where does the r' come from?

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0Oh! so another way to look at d^2r/dt^2 is r''?

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0Okay! 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?

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1well 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

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0So, since r(1)=1, then t(1)=2, then r of t = 1? Somehow I confused myself...

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1I'm sorry to be so blunt

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0No, it makes no sense to me either.

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1You are given r'(1)=1 and r(1)=1

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1r'(1)=1 says when t=1 r'=1

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1\[1=\frac{1}{1^2}+C_1 \] The reason why I wrote this

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1I replaced t with 1 and I replaced r' with 1

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0That makes sense...what happens to the 2 of t(1)=2 tho? how would I solve for C then? Or is it just replace r' with 1 and t with 1?

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0Sorry for all these questions...

freckles
 3 years ago
Best ResponseYou've already chosen the best response.1I don;t understand where you are seeing t(1)=2?

sharathe
 3 years ago
Best ResponseYou've already chosen the best response.0I added it after getting the orignal problem in the "what happens after" part...cause the next problem was basically this problem, but instead of t(1)=1, it was t(2)=1...Sorry if this was all confusing! I don't quite get how to solve for C when the r' and t equal different numbers but have the same number plugged in...

Chlorophyll
 3 years ago
Best ResponseYou've already chosen the best response.2@sharathe you're confused between the input t = 1, output r' ( 1) = 1 From these input and output => constant C1 = 2
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