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]
