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Question Number 68825 by peter frank last updated on 15/Sep/19

Answered by mind is power last updated on 16/Sep/19

$$dx(1+e^{\frac{x}{y}})+e^{\frac{x}{y}}(1-\frac{x}{y})dy$$$$p=1+e^{\frac{x}{y}}$$$$q=e^{\frac{x}{y}}(1-\frac{x}{y})$$$$\frac{\partial p}{\partial y}=-x\frac{e^{\frac{x}{y}}}{y^2}$$$$\frac{\partial q}{\partial x}=\frac{e^{\frac{x}{y}}}{y}(1-\frac{x}{y})-\frac{e^{\frac{x}{y}}}{y}=-x\frac{e^{\frac{x}{y}}}{y^2}=\frac{\partial p}{\partial y}$$$$soexactequation$$

Commented by peter frank last updated on 22/Sep/19

$$thankyou$$

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