T_TEngineer_AdamT_T
New member
- Joined
- Apr 15, 2007
- Messages
- 24
obtain the particular solution of
\(\displaystyle 2ydx = 3xdy\) x = 2, y = -1
\(\displaystyle \frac{2dx}{x} = \frac{3dy}{y}\)
\(\displaystyle 2lnx = 3lny + C\)
\(\displaystyle 2\ln{2} = 3\ln(-1) + C\) <<< im stuck right here... there's suppose to be no negative ln right?
\(\displaystyle 2ydx = 3xdy\) x = 2, y = -1
\(\displaystyle \frac{2dx}{x} = \frac{3dy}{y}\)
\(\displaystyle 2lnx = 3lny + C\)
\(\displaystyle 2\ln{2} = 3\ln(-1) + C\) <<< im stuck right here... there's suppose to be no negative ln right?