अवकल समीकरण `(dy)/(dx)+(y)/(x)=y^(2)` का हल क्या है?

Options

A. `x^(2)ylog_(e)(C//x)=1`
B. `xy^(2)log_(e)(C//x)=1`
C. `xylog_(e)(C//x^(2))=1`
D. `xylog_(e)(C//x)=1`

Answer

D

Explanation

दिया है, `(dy)/(dx)+(y)/(x)=y^(2)implies(1)/(y^(2))(dy)/(dx)+(1)/(yx)=1`
माना `v=(1)/(y)implies(dv)/(dx)=-(1)/(y^(2))(dy)/(dx)implies-(dv)/(dx)+(v)/(x)=1implies(dv)/(dx)-(v)/(x)=-1(dv)/(dx)+Pv=Q` से तुलना करने पर, `P=-1//x, Q=-1`
`IF=e^(intPdx)=e^(int-(dx)/(x))=+1//x`
`therefore v.(IF)=intQ(IF)dx+C_(1)`
`implies (v)/(x)=int-1.(1)/(x)dx+C_(1)=-log_(e)x+C_(1)`
`implies (1)/(xy)=-log_(e)x+C_(1)" "[because C_(1)=log_(e)C]`
`implies (1)/(xy)=-log_(e)x+log_(e)C=log_(e)((C)/(x))impliesxy log_(e)((C)/(x))=1`