Question

Show that the general solution of the differential equation is given by ( x + y + 1) = A (1 – x – y – 2 xy ), where A is parameter

Answer 1

The differential equation is dy dx + y 2 +y+1 x 2 +x+1 =0.

Simplify given equation.

dy dx + y 2 +y+1 x 2 +x+1 =0 dy dx =− { y 2 +y+1 } { x 2 +x+1 } dy ( y 2 +y+1 ) =− dx ( x 2 +x+1 ) dy ( y 2 +y+1 ) + dx ( x 2 +x+1 ) =0

Integrating both sides of the above equation,

∫ dy ( y 2 +y+1 ) + { ∫ dx ( x 2 +x+1 ) }=C ∫ dy ( y+ 1 2 ) 2 + ( 3 2 ) 2 + ∫ dx ( x+ 1 2 ) 2 + ( 3 2 ) 2 =C ( 2 3 ) tan −1 { 2y+1 3 }+( 2 3 ) tan −1 { 2x+1 3 }=C tan −1 [ 2y+1 3 + 2x+1 3 1−{ 2y+1 3 }×{ 2x+1 3 } ]= 3 C 2

Further simplify the above equation.

tan −1 [ 2x+2y+2 3 1−{ 4xy+2x+2y+1 3 } ]= 3 C 2 tan −1 [ { 2 3 ( x+y+1 ) } { 3−4xy−2x−2y−1 } ]= 3 C 2 tan −1 [ 3 ( x+y+1 ) 2( 1−x−y−2xy ) ]= 3 C 2 3 ( x+y+1 ) 2( 1−x−y−2xy ) =tan( 3 C 2 )

Consider, B=tan( 3 C 2 )and simplify the above equation.

3 ( x+y+1 ) 2( 1−x−y−2xy ) =B ( x+y+1 )= 2B 3 ( 1−x−y−2xy ) ( x+y+1 )=A( 1−x−y−2xy ) [ A= 2B 3 ]

Hence, it is proved that ( x+y+1 )=A( 1−x−y−2xy ).