S sraz New member Joined Aug 19, 2009 Messages 2 Aug 19, 2009 #1 dy/dx = (ycosx) / (1 + y^2) ; y(0) = 1 I'm confused as to where to go with this one. So far I've been able to get to this... ln|y| + [y^2]/2 = sinx I'm lost at this point. Sorry if I'm doing this completely wrong, any help is appreciated.
dy/dx = (ycosx) / (1 + y^2) ; y(0) = 1 I'm confused as to where to go with this one. So far I've been able to get to this... ln|y| + [y^2]/2 = sinx I'm lost at this point. Sorry if I'm doing this completely wrong, any help is appreciated.
G galactus Super Moderator Staff member Joined Sep 28, 2005 Messages 7,203 Aug 19, 2009 #2 You have the worst part done. Just let y=1 and x=0 and solve for c. It is difficult to solve this for y, so just leave it in the form you have after you find c. \(\displaystyle ln(y)+\frac{y^{2}}{2}=sin(x)+C\)
You have the worst part done. Just let y=1 and x=0 and solve for c. It is difficult to solve this for y, so just leave it in the form you have after you find c. \(\displaystyle ln(y)+\frac{y^{2}}{2}=sin(x)+C\)
