Question

How would I solve the following differential equation?
y'=(t-y)^2

Answer 1

burnoodle it

Answer 2

maybe

Answer 3

y'=(t-y)^2

y'=t^2-2ty+y^2

hmm

or try the wolfs method : http://www.wolframalpha.com/input/?i=y%27%3D%28t-y%29%5E2

Answer 4

they appear to just sub in a new function of t: v=(t-y)

Answer 5

that's pretty dope

Answer 6

v' = 1-y' ; y' = -v'+1


-v'+1 = v^2

Answer 7

v' = -v^2+1
\[\int (\frac{v'}{-v^2+1}=1)\]
\[-\frac{1}{2}\int \frac{-2v'}{-v^2+1}dt=t+c\]
i think i lose it about there