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प्रश्न
Find the particular solution of differential equation:
`dy/dx=-(x+ycosx)/(1+sinx) " given that " y= 1 " when "x = 0`
उत्तर
`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))`
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