Question

Find a particular solution of the differential equation , given that y = 0 when x = 0

Answer 1

The given differential equation is,

( x+1 ) dy dx =2 e −y −1

The above equation can be written as,

dy 2 e −y −1 = dx x+1

Integrate both sides of equation.

∫ dy 2 e −y −1 = ∫ dx x+1 ∫ dy 2 e y −1 =log| x+1 |+c ∫ e y dy 2− e y =log| x+1 |+c

Consider,

t=2− e y −dt= e y dy

Now, integral becomes,

∫ −dt t =log| x+1 |+c −logt=log| x+1 |+c −log( 2− e y )=log| x+1 |+c (1)

Given that y=0 when x=0.

Substitute the values in the equation (1),

−log( 2− e 0 )=log( 0+1 )+c −log1=log1+c c=0

Substitute the value of c in the equation (1). So, the solution of the differential equation is,

−log( 2− e y )=log( x+1 )+0 log( 2− e y )=log( 1 x+1 ) 2− e y = 1 x+1 e y = 2x+1 x+1

Taking log both sides,

y=log| 2x+1 x+1 |       [ x≠−1 ]

Thus, the solution of the differential equation is y=log| 2x+1 x+1 |.