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Question Number 171442 by mnjuly1970 last updated on 15/Jun/22

Commented by infinityaction last updated on 15/Jun/22

$$e^x(1+x)dx+({ye}^y-{xe}^x)dy=0$$$$lett={xe}^x$$$$e^x(1+x)dx=dt$$$$dt+({ye}^y-t)dy=0$$$$\frac{dt}{dy}+{ye}^y-t=0$$$$\frac{dt}{dy}-t=-{ye}^y$$$$\frac{dt}{dy}+Pt=Q$$$$P=-1,Q=-{ye}^y$$$$I.F=e^{\int -1dy}\Rightarrow I.F=e^{-y}$$$$solution$$$$t\times I.F=\int (I.F\times Q)dy+C$$$${te}^{-y}=\int (-{ye}^y\times e^{-y})dy+C$$$${te}^{-y}+\frac{y^2}{2}=C$$$$\therefore t={xe}^x$$$${xe}^{(x-y)}+\frac{y^2}{2}=C$$

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