Answer to Question #187019 in Differential Equations for Eeman Zafar

Question #187019

(π’š 𝟐 + π’™π’šπŸ‘ ) 𝒅𝒙 + (πŸ“π’š 𝟐 βˆ’ π’™π’š + π’š 𝟐 𝐬𝐒𝐧 π’š) π’…π’š = 𝟎 


1
Expert's answer
2021-05-07T10:28:52-0400

Given equation is of the form- Mdx+Ndy=0Mdx+Ndy=0


where, M(x,y)=y2+xy3β‡’My=2y+3xy2M(x,y)=y^2+xy^3\Rightarrow M_y=2y+3xy^2

N(x,y)=5y2βˆ’xy+y2sinyβ‡’Nx=βˆ’yN(x,y)=5y^2-xy+y^2siny\Rightarrow N_x=-y


So, Nxβˆ’MyM=βˆ’yβˆ’(2y+3xy2)y2+xy3=βˆ’3y(1+xy)y2(1+y)=βˆ’3y\dfrac{N_x-M_y}{M}=\dfrac{-y-(2y+3xy^2)}{y^2+xy^3}=\dfrac{-3y(1+xy)}{y^2(1+y)}=\dfrac{-3}{y}



Integrating Factor ΞΌ=eβˆ’βˆ«3ydy=eβˆ’3lny=yβˆ’3\mu=e^{-\int\frac{3}{y}dy}=e^{-3lny}=y^{-3}


So Its, solution is-


(1y+x)dx+(5yβˆ’xy2+siny)dy=0(\dfrac{1}{y}+x)dx+(\dfrac{5}{y}-\dfrac{x}{y^2}+siny)dy=0


fx=1y+x,fy=5yβˆ’xy2+sinyf_x=\dfrac{1}{y}+x, f_y=\dfrac{5}{y}-\dfrac{x}{y^2}+siny


and ,

F(x,y)=Fx+Fy=x22+xy+5lnyβˆ’cosyF(x,y)=F_x+F_y=\dfrac{x^2}{2}+\dfrac{x}{y}+5lny-cosy


So The complete solution is-

F(x,y)=cF(x,y)=c


β‡’x22+xy+5lnyβˆ’cosy=c\Rightarrow \dfrac{x^2}{2}+\dfrac{x}{y}+5lny-cosy=c


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