Question

Find the particular solution of the differential equation , given that y = 1 when x = 0

Answer 1

The differential equation is ( 1+ e 2x )dy+( 1+ y 2 ) e x dx=0, y=1 when x=0.

Simplify above equation.

( 1+ e 2x )dy+( 1+ y 2 ) e x dx=0 ( dy 1+ y 2 )+ e x dx ( 1+ e 2x ) =0

By integrating both sides of the above equation we get,

tan −1 y+ ∫ e x dx ( 1+ e 2x ) =C(1)

Let, e x =t.

Differentiate with respect to x.

e x =t e 2x = t 2 d dx ( e x )= dt dx e x dx=dt

Substitute the above value in the equation (1)d.

tan −1 y+ ∫ dt 1+ t 2 =C tan −1 y+ tan −1 t=C tan −1 y+ tan −1 ( e x )=C (2)

Now, substitute y=1 and x=0.

tan −1 1+ tan −1 ( 1 )=C { π 4 }+{ π 4 }=C π 2 =C

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

tan −1 y+ tan −1 ( e x )= π 2

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