# Find the particular solution of differential equation: dy/dx=(−x+ycosx)/(1+sinx) given that y=1 when x=0 - Mathematics

Find the particular solution of differential equation:

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

#### Solution

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))

Concept: General and Particular Solutions of a Differential Equation
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