Find all possible functions

dy = (1/t^2)dt
y=-(1/t) + C
C can be anything, t can be anything except 0.
 
Gene said:
dy = (1/t^2)dt
y=-(1/t) + C
C can be anything, t can be anything except 0.
Click to expand...

This one I really don't understand:

y'= -1/1 + c

if t = 1 and c =1 then -1 + 1 = 0

if t=2 and c = 2 then -1/2 + 2 = 1.5
 
If y = -1/t + C then
y' = dy/dt = d(-1)*t^-1) + d(C)=
-1*d(t^-1) + 0 =
t^-2 =
1/t^2 =
1/square of t as required.

I don't understand your objection to y=0 or y=1.5. What am I missing?
Note: It is y=-1/t+C not y'=-1/t+C
 
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