Question

Find a particular solution of the differential equation , given that y = – 1, when x = 0 (Hint: put x – y = t )

Answer 1

Let the differential equation is ( x−y )( dx+dy )=dx−dy, y=−1 when x=0.

Simplify above equation.

( x−y )( dx+dy )=dx−dy ( x−y+1 )dy=( 1−x+y )dx dy dx = ( 1−x+y ) ( x−y+1 ) dy dx = 1−( x−y ) 1+( x−y ) (1)

Let ( x−y )=t and differentiate with respect to x.

d dx ( x−y )= dt dx 1− dy dx = dt dx 1− dt dx = dy dx

Substitute the value of x−y and dy dx in the equation (1).

1− dt dx = 1−t 1+t dt dx =1− 1−t 1+t dt dx = ( 1+t )−( 1−t ) ( 1+t ) dt dx = 2t ( 1+t )

Further simplify the above equation.

( 1+t t )dt=2dx ( 1+ 1 t )dt=2dx

By integrating both side of the above equation, we get

t+log| t |=2x+C ( x−y )+log| x−y |=2x+C (2)

Now, substitute y=−1 and x=0.

log1=0−1+C C=1

Substitute the value of C=1in the equation (2).

( x−y )+log| x−y |=2x+1 log| x−y |=x+y+1

Thus, the above equation is required solution for differential equation.