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=0"


where, "M(x,y)=y^2+xy^3\\Rightarrow M_y=2y+3xy^2"

"N(x,y)=5y^2-xy+y^2siny\\Rightarrow N_x=-y"


So, "\\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 "\\mu=e^{-\\int\\frac{3}{y}dy}=e^{-3lny}=y^{-3}"


So Its, solution is-


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


"f_x=\\dfrac{1}{y}+x, f_y=\\dfrac{5}{y}-\\dfrac{x}{y^2}+siny"


and ,

"F(x,y)=F_x+F_y=\\dfrac{x^2}{2}+\\dfrac{x}{y}+5lny-cosy"


So The complete solution is-

"F(x,y)=c"


"\\Rightarrow \\dfrac{x^2}{2}+\\dfrac{x}{y}+5lny-cosy=c"


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