Question

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

C=

Answer 1

`y(1+x)+x-5\ln xy=C`

Explanation:

Consider the given differential equation is

`(1-(5)/(y)+x)(dy)/(dx)+y=(5)/(x)-1`

`(1-(5)/(y)+x)(dy)/(dx)=(5)/(x)-1-y`

`(1-(5)/(y)+x)dy=((5)/(x)-1-y)dx`

Taking all variables on right sides.

`(1-(5)/(y)+x)dy-((5)/(x)-1-y)dx=0`

`(-(5)/(x)+1+y)dx+(1-(5)/(y)+x)dy=0`

Let as assume,

`M=-(5)/(x)+1+y` and
`N=1-(5)/(y)+x`

Find partial derivatives
`(\partial M)/(\partial y)` and
`(\partial N)/(\partial x)`

`(\partial M)/(\partial y)=1` and
`(\partial N)/(\partial x)=1`

Since
`(\partial M)/(\partial y)=(\partial N)/(\partial x)`, therefore the given differential equation is exact.

The solution of the exact differential equation is

`\int Mdx+\int N(\text{without x)}dy=C`

`\int (-(5)/(x)+1+y)dx+\int (1-(5)/(y))dy=C`

`yx-5\ln x+x+y-5\ln y=C`

`y+x+xy-5\ln x-5\ln y=C`

`y(1+x)+x-5(\ln x+\ln y)=C`

`y(1+x)+x-5\ln xy=C`