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- Jan 26, 2012

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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.

- Jan 26, 2012

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If it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$.

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- Jan 26, 2012

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Nice! The result is separable. I getIf it is indeed what Ackbach says, try letting $u = \dfrac{x+y}{2x-y+1}$.

$$\frac{e^{-u}}{u+1}\,u'=\frac{1}{3x+1}.$$

Of course, the integral on the left is not elementary. Oh, well.

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