हल कीजिए-
`y-x(dy)/(dx)=a(y^(2)+(dy)/(dx))`

Explanation

`y-x (dy)/(dx) =a (y^(2)+(dy)/(dx))`
`impliesy-x (dy)/(dx) =ay^(2)+a(dy)/(dx)`
`implies(y-ay^(2))=(x+a)(dy)/(dx)`
`implies (y-ay^(2))dx=(x+a) (dy)/(dx)`
`implies (y-ay^(2))dx=(x+a)dy`
`implies(dy)/(x+a)=(dy)/(y(1-ay))`
समाकलन करने पर,
`int (dx)/(x+a)=int (dy)/(y(1- ay))" "...(1)`
अब `(1)/(y (1- ay))=A/y+(B)/(1-ay)" "` [आंशिक भिन्न से]
`1=A(1-ay)+By" "...(2)`
समी (2 ) में, `y =0 ` रखने पर `A =1 `
समी (2 ) में `1-ay=0` अर्थात `y=1/a` रखने पर `B =a `
`therefore (1)/(y(1-ay))=(1)/(y) +(a)/(1-ay)" "...(3)`
समी (1 ) और (3 ) से,
`int (1)/(1+x)dx=int ((1)/(y) +(a)/(a-ay))dy`
` impliesint (1)/(a+x)dx=int 1/y dy+a int (1)/(1-ay)dy`
माना `1-ay =timplies -ady =dt=-(dt)/(a)`
`therefore int (1)/(a+x)dx =int 1/y -int(dt)/(t)`
`implieslog |a+x|=log y -log |t|+ log C `
`implieslog |a+x|=log |y|-log |1-ay|+ log C`
`implieslog |a+x|+log |1-ay|-log |y|=log C`
`implieslog |(a+x)(1-ay)|-log | y|=log C`
`implieslog ""((1+x)(1-ay))/(|y|)=log C`
`implies((a+x)(1-ay))/(y)=C`
`implies (x+a)(1-ay)=Cy`