In April 2025
Answer to Question #291549 in Differential Equations for chris
(y - xy)dx + xdy = 0
"M=(y-xy)\\\\\n\\frac{dM}{dy}=1-xy\\\\\nN= x\\\\\n\\frac{dN}{dy}=1\\\\\nwhich ~is~ not~exact\\\\\n( \\frac{dM}{dy}-\\frac{dN}{dy})=f(x)\\\\\n(1-x-1)=f(x),-x=f(x)\\\\\n\\frac{1}{N}( \\frac{dM}{dy}-\\frac{dN}{dy})=f(x)\\\\\n\\frac{1}{x}(-x)=f(x)\\\\\n-1=f(x)\\\\\nfind~the~integrating~factor\\\\\n\\frac{1}{N}( \\frac{dM}{dy}-\\frac{dN}{dy})=f(x), I.F=e^{\\int f(x)dx}\\\\\n\\frac{1}{M}( \\frac{dM}{dy}-\\frac{dN}{dy})=f(y), I.F=e^{\\int -f(y)dx}\\\\\n I.F=e^{\\int f(x)dx}=e^{\\int -1dx}=e^{-x}\\\\\ne^{-x}[(y-xy)dx+xdy=0]\\\\\n(ye^{-x}-xye^{-x})dx+xe^{-x}dy=0\\\\\ntest~for~exactness\\\\\n\\frac{dM}{dy}=e^{-x}-xe^{-x},\\frac{dN}{dy}=e^{-x}-xe^{-x}\\\\\n\\frac{dF}{dx}=M(x,y)\\\\\n\\int dF=\\int (ye^{-x}-xye^{-x})dx\\\\\nF=-ye^{-x}-y\\int xe^{-x}dx\\\\\nwe~use~integration~by~part~to~solve~\\int xe^{-x}dx\\\\\n\\int xe^{-x}dx=xe^{-x}+e^{-x}\\\\\nF=-ye^{-x}+yxe^{-x}+ye^{-x}+Q(y)\\\\\nF=xye^{-x}+Q(y)\\\\\n\\frac{dF}{dy}=N(x,y)\\\\\n\\int dF=\\int (xe^{-x})dy\\\\\nF=(xe^{-x})\\int dy\\\\\nF=xye^{-x}+R(x),then\\\\\nF=xye^{-x}+Q(y)=xye^{-x}+R(x)\\\\Q(x)=R(x)=0\\\\\nxye^{-x}+Q(y)=F\\\\\nxye^{-x}=c\\\\\nThe~general~equation"
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