Answer to Question #291549 in Differential Equations for chris

Question #291549

(y - xy)dx + xdy = 0



1
Expert's answer
2022-01-31T09:36:25-0500

M=(yxy)dMdy=1xyN=xdNdy=1which is not exact(dMdydNdy)=f(x)(1x1)=f(x),x=f(x)1N(dMdydNdy)=f(x)1x(x)=f(x)1=f(x)find the integrating factor1N(dMdydNdy)=f(x),I.F=ef(x)dx1M(dMdydNdy)=f(y),I.F=ef(y)dxI.F=ef(x)dx=e1dx=exex[(yxy)dx+xdy=0](yexxyex)dx+xexdy=0test for exactnessdMdy=exxex,dNdy=exxexdFdx=M(x,y)dF=(yexxyex)dxF=yexyxexdxwe use integration by part to solve xexdxxexdx=xex+exF=yex+yxex+yex+Q(y)F=xyex+Q(y)dFdy=N(x,y)dF=(xex)dyF=(xex)dyF=xyex+R(x),thenF=xyex+Q(y)=xyex+R(x)Q(x)=R(x)=0xyex+Q(y)=Fxyex=cThe general equationM=(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


Need a fast expert's response?

Submit order

and get a quick answer at the best price

for any assignment or question with DETAILED EXPLANATIONS!

Comments

No comments. Be the first!

Leave a comment