Dec 25, 2012 Thread starter #1 H hatguy New member Dec 25, 2012 3 I need to solve the following ODE: but i can't figure out a way to. Please help! Last edited by a moderator: Dec 25, 2012
Dec 25, 2012 Admin #2 Ackbach Indicium Physicus Staff member Jan 26, 2012 4,205 Regarding 3: Did you mean $$y'=e^{ \frac{x+y}{2x-y+1}}+\frac{3y-1}{3x+1}?$$ The absence of a closing parenthesis in the numerator of the argument of the exponential function makes your meaning unclear.
Regarding 3: Did you mean $$y'=e^{ \frac{x+y}{2x-y+1}}+\frac{3y-1}{3x+1}?$$ The absence of a closing parenthesis in the numerator of the argument of the exponential function makes your meaning unclear.
Dec 25, 2012 #3 Jester Well-known member MHB Math Helper Jan 26, 2012 183 If it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$.
Dec 27, 2012 Admin #4 Ackbach Indicium Physicus Staff member Jan 26, 2012 4,205 Jester said: If it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$. Click to expand... Nice! The result is separable. I get $$\frac{e^{-u}}{u+1}\,u'=\frac{1}{3x+1}.$$ Of course, the integral on the left is not elementary. Oh, well. Last edited: Dec 27, 2012
Jester said: If it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$. Click to expand... Nice! The result is separable. I get $$\frac{e^{-u}}{u+1}\,u'=\frac{1}{3x+1}.$$ Of course, the integral on the left is not elementary. Oh, well.