Another math problem. I hope some of you can do this one.
Find the orthogonal trajectories of y=ln(tanx+c)
this is what i did. e^y=tanx+c, e^y(dy/dx)=(secx)^2
(dy/dx)=((secx)^2)/e^y, Now set (dy/dx)=-(e^y)[(cosx)^2]
so (-e^-y)dy=[(cosx)^2]dx
answer is
e^-y=(sinxcosx+x)/2 + K
Did i do it right.
Find the orthogonal trajectories of y=ln(tanx+c)
this is what i did. e^y=tanx+c, e^y(dy/dx)=(secx)^2
(dy/dx)=((secx)^2)/e^y, Now set (dy/dx)=-(e^y)[(cosx)^2]
so (-e^-y)dy=[(cosx)^2]dx
answer is
e^-y=(sinxcosx+x)/2 + K
Did i do it right.