- #1
doroulla
- 16
- 0
hi. I have asked for help on this again but i got even more confused. The question says solve:
dx/dt - 2t(2x-1) = 0 when t=0 when x=0.
I did:
0.5 ln(2x-1) = 2t^2 / 2 + K
then i have 2x-1 = 0.5(e^t^2) + K
putting x=0 and t=0 i get:
K=0
thus i have 2x-1 = e^(2t^2)
thus: x=0.5e^(2t^2) + 1
but i know it is wrong because when i put t=0 i do not get x=0
someone please help me. I've been stuck on this coursework for a week! Thank you
dx/dt - 2t(2x-1) = 0 when t=0 when x=0.
I did:
0.5 ln(2x-1) = 2t^2 / 2 + K
then i have 2x-1 = 0.5(e^t^2) + K
putting x=0 and t=0 i get:
K=0
thus i have 2x-1 = e^(2t^2)
thus: x=0.5e^(2t^2) + 1
but i know it is wrong because when i put t=0 i do not get x=0
someone please help me. I've been stuck on this coursework for a week! Thank you