The solution of `(x^2+x y)dy=(x^2+y^2)dx` is (a) `( b ) (c)logx=log(( d ) (e) x-y (f))+( g ) y/( h ) x (i) (j)+c (k)` (l) (m) `( n ) (o)logx=2log(( p ) (q) x-y (r))+( s ) y/( t ) x (u) (v)+c (w)` (x) (y) `( z ) (aa)logx=log(( b b ) (cc) x-y (dd))+( e e ) x/( f f ) y (gg) (hh)+c (ii)` (jj) (kk) None of these

Options

A. `logx = log(x-y)+y/x+c`
B. `logx = 2log(x-y)+y/x+c`
C. `logx= log(x-y)+x/y+c`
D. None of these

Answer

B

Explanation

`(x^(2)+xy)dy=(x^(2)+y^(2))dx` or `(dy)/(dx) = (x^(2)+y^(2))/(x^(2)+xy)`
Thus, equation reduces to
`x(dv)/(dx) = (1+v^(2))/(1+v) -v=(1+v^(2)-v-v^(2))/(1+v)=(1-v)/(1+v)`
or `-int(1-2/(1-v))dv=int(dx)/x`
or `-v-2log(1-v)=logx+logc`
or `-y/x-2log(x-y)/x=logx+logc`
or `logx=2log(x-y)+y/x+k` where `k=logc`