Solve the following differential equation: (1− 5 y +x) dy/dx +y= 5/x −1 .C=

Answer:[tex]y(1+x)+x-5\ln xy=C[/tex]Step-by-step explanation:Consider the given differential equation is[tex](1-\frac{5}{y}+x)\frac{dy}{dx}+y=\frac{5}{x}-1[/tex][tex](1-\frac{5}{y}+x)\frac{dy}{dx}=\frac{5}{x}-1-y[/tex][tex](1-\frac{5}{y}+x)dy=(\frac{5}{x}-1-y)dx[/tex]Taking all variables on right sides.[tex](1-\frac{5}{y}+x)dy-(\frac{5}{x}-1-y)dx=0[/tex][tex](-\frac{5}{x}+1+y)dx+(1-\frac{5}{y}+x)dy=0[/tex]Let as assume,[tex]M=-\frac{5}{x}+1+y[/tex] and [tex]N=1-\frac{5}{y}+x[/tex]Find partial derivatives [tex]\frac{\partial M}{\partial y}[/tex] and [tex]\frac{\partial N}{\partial x}[/tex][tex]\frac{\partial M}{\partial y}=1[/tex] and [tex]\frac{\partial N}{\partial x}=1[/tex]Since [tex]\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}[/tex], therefore the given differential equation is exact.The solution of the exact differential equation is[tex]\int Mdx+\int N(\text{without x)}dy=C[/tex][tex]\int (-\frac{5}{x}+1+y)dx+\int (1-\frac{5}{y})dy=C[/tex][tex]yx-5\ln x+x+y-5\ln y=C[/tex][tex]y+x+xy-5\ln x-5\ln y=C[/tex][tex]y(1+x)+x-5(\ln x+\ln y)=C[/tex][tex]y(1+x)+x-5\ln xy=C[/tex]