Answer to Question #291549 in Differential Equations for chris

Question

(y - xy)dx + xdy = 0

Answer 1

$M=(y-xy)\\ \frac{dM}{dy}=1-xy\\ N= x\\ \frac{dN}{dy}=1\\ which ~is~ not~exact\\ ( \frac{dM}{dy}-\frac{dN}{dy})=f(x)\\ (1-x-1)=f(x),-x=f(x)\\ \frac{1}{N}( \frac{dM}{dy}-\frac{dN}{dy})=f(x)\\ \frac{1}{x}(-x)=f(x)\\ -1=f(x)\\ find~the~integrating~factor\\ \frac{1}{N}( \frac{dM}{dy}-\frac{dN}{dy})=f(x), I.F=e^{\int f(x)dx}\\ \frac{1}{M}( \frac{dM}{dy}-\frac{dN}{dy})=f(y), I.F=e^{\int -f(y)dx}\\ I.F=e^{\int f(x)dx}=e^{\int -1dx}=e^{-x}\\ e^{-x}[(y-xy)dx+xdy=0]\\ (ye^{-x}-xye^{-x})dx+xe^{-x}dy=0\\ test~for~exactness\\ \frac{dM}{dy}=e^{-x}-xe^{-x},\frac{dN}{dy}=e^{-x}-xe^{-x}\\ \frac{dF}{dx}=M(x,y)\\ \int dF=\int (ye^{-x}-xye^{-x})dx\\ F=-ye^{-x}-y\int xe^{-x}dx\\ we~use~integration~by~part~to~solve~\int xe^{-x}dx\\ \int xe^{-x}dx=xe^{-x}+e^{-x}\\ F=-ye^{-x}+yxe^{-x}+ye^{-x}+Q(y)\\ F=xye^{-x}+Q(y)\\ \frac{dF}{dy}=N(x,y)\\ \int dF=\int (xe^{-x})dy\\ F=(xe^{-x})\int dy\\ F=xye^{-x}+R(x),then\\ F=xye^{-x}+Q(y)=xye^{-x}+R(x)\\Q(x)=R(x)=0\\ xye^{-x}+Q(y)=F\\ xye^{-x}=c\\ The~general~equation$