Question

Find the particular solution of differential equation:

`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`

Answer 1

`dy/dx=-(x+ycosx)/(1+sinx)`

⇒ `dy/dx+cosx/(1+sinx)y=x/(1+sinx )" ......i"`

This is a linear differential equation with

`P=cosx/(1+sinx),Q =-x/(1+sinx)`

`:.I.F. = e^intcosx/(1+sinx)dx`

= `e^log(1+sinx)`

= 1+ sinx

Multiplying both the sides of i by I.F. = 1 + sinx, we get

`(1+sinx)dy/dx+ycosx=-x`

Integrating with respect to x, we get

`y(1+sinx)=int-xdx+C`

`=>y =(2C-x^2)/(2(1+sinx)) " ....(ii)"`

Given that y = 1 when x = 0

`:.1=(2C)/(2(1+0))`

⇒ C =1 ................(iii)

Put iii in ii , we get

`y = (2-x^2)/(2(1+sinx))`