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प्रश्न
Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0
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उत्तर
The given differential equation is `"dy"/"dx" = "xy"/("x"^2+"y"^2)` ....(i)
Let y = vx, then .......(ii)
`"dy"/"dx"="v" + "x" "dv"/"dx"` ....(iii)
from (i), (ii) and (iii), we get
`"v" + "x" "dv"/"dx" = "vx"^2/("x"^2+"v"^2"x"^2)`
`⇒ "v" + "x" "dv"/"dx" = "v"^2/(1+"v"^2)`
`⇒ "x" "dv"/"dx" = ("v"^2-"v"-"v"^3)/(1+"v"^2)`
`⇒((1+"v"^2))/("v"^3+"v"-"v"^2)"dv" = -"dx"/"x"`
`⇒(("v"^2 +1-"v"+"v"))/("v"("v"^2+1-"v"))"dv" = -"dx"/"x"`
`⇒(1/"v" + 1/("v"^2+1-"v"))"dv" = -"dx"/"x"`
Now, Integrate both the sides
`⇒ int (1/"v"+1/("v"^2+1-"v"))"dv" = - int "dx"/"x"`
`⇒ int1/"v""dv" + int1/("v"^2+1-"v")"dv" = - int"dx"/"x"`
`⇒int 1/"v" "dv" +int 1/("v"^2-2."v". 1/2+1/4+1-1/4)"dv" = -int"dx"/"x"`
`⇒int 1/"v" "dv" + int 1/(("v"-1/2)^2+(sqrt3/2)^2) "dv" = -int"dx"/"x"`
`⇒ "lnv" + 2/sqrt3 tan^-1((2"v"-1)/sqrt3)= - "lnx"+"c"`
`⇒"lny" + 2/sqrt3 tan^-1((2"y"-"x")/(sqrt3"x"))="c"`
It is given that y = 1 when x = 0,
Therefore c =`pi/sqrt3`
Hence, the particular solution of the given differential equation is `"ln y" + 2/sqrt3 tan^-1((2"y"-"x")/(sqrt3"x")) = pi/sqrt3.`
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