Question

For the differential equation find the solution curve passing through the point (1, –1).

Answer 1

The differential equation of the curve is given as,

xy dy dx =( x+2 )( y+2 )

Simplify differential equation,

y ( y+2 ) dy dx = ( x+2 ) x y ( y+2 ) dy= ( x+2 ) x dx

Integrate both sides.

∫ y y+2 dy = ∫ x+2 x dx ∫ ( 1− 2 y+2 ) dy= ∫ ( 1+ 2 x ) dx ∫ dy − ∫ 2 y+2 dy = ∫ dx + ∫ 2 x dx y−2log| y+2 |=x+2log| x |+C

Simplify,

y−x−C=2( logx+log( y+2 ) ) y−x−C=log( x 2 ( y+2 ) 2 ) (1)

Since this curve passes through point ( 1,−1 ), so it will satisfy the above curve,

−1−1−C=log( 1 ( −1+2 ) 2 ) −2−C=0 C=−2

Substitute the value of Cinto equation (1),

y−x+2=log[ x 2 ( y+2 ) 2 ]

Thus, the required equation of the curve that passes through point ( 1,−1 ) is y−x+2=log[ x 2 ( y+2 ) 2 ].