The solution (x+y+1) dy =dx are-

Options

A. `x+y+2=Ce^(y)`
B. `x+y+4=C log y`
C. `log (x+y+2)=Cy`
D. none of these

Answer

A

Explanation

`(dy)/(dx)=(1)/(x+y+1)`
`(dv)/(dx)-1=(1)/(v)`
`(dv)/(dx)=(1)/(v)+1`
`(v)/(v+1)dv=dx`
` Put" " v =x+y+1`
`(dv)/(dx)=1+(dy)/(dx)`
`(1-(1)/(v+1))dv=dx` integrating both side , we get
`int(1-(1)/(v+1))dv=int dx`
`v-log(v+1)=x+c`
` x+y+1-log(x+y+1div1)=x+c`
`y+1-C=log (x+y+2)`
`x+y +2= e^(y+1-c)`
` =e^(y),e^(1-c)`
`=e^(y),C`