Find the particular solution of the differential equation "dy"/"dx" = "xy"/("x"^2+"y"^2),given that y = 1 when x = 0

प्रश्न

Find the particular solution of the differential equation `"dy"/"dx" = "xy"/("x"^2+"y"^2),`given that y = 1 when x = 0

योग

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 $v + x \frac{d v}{d x} = \frac{v x^{2}}{x^{2} + v^{2} x^{2}} \Rightarrow v + x \frac{d v}{d x} = \frac{v^{2}}{1 + v^{2}} \Rightarrow x \frac{d v}{d x} = \frac{v^{2} - v - v^{3}}{1 + v^{2}} \Rightarrow \frac{(1 + v^{2})}{v^{3} + v - v^{2}} d v = - \frac{d x}{x} \Rightarrow \frac{(v^{2} + 1 - v + v)}{v (v^{2} + 1 - v)} d v = - \frac{d x}{x} \Rightarrow (\frac{1}{v} + \frac{1}{v^{2} + 1 - v}) d v = - \frac{d x}{x}$

`"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 $\Rightarrow \int (\frac{1}{v} + \frac{1}{v^{2} + 1 - v}) d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} + 1 - v} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} - 2 \cdot v \cdot \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{{(v - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d v = - \int \frac{d x}{x} \Rightarrow \ln v + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 v - 1}{\sqrt{3}}) = - \ln x + c \Rightarrow \ln y + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - x}{\sqrt{3} x}) = c$ $\Rightarrow \int (\frac{1}{v} + \frac{1}{v^{2} + 1 - v}) d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} + 1 - v} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{v^{2} - 2 \cdot v \cdot \frac{1}{2} + \frac{1}{4} + 1 - \frac{1}{4}} d v = - \int \frac{d x}{x} \Rightarrow \int \frac{1}{v} d v + \int \frac{1}{{(v - \frac{1}{2})}^{2} + {(\frac{\sqrt{3}}{2})}^{2}} d v = - \int \frac{d x}{x} \Rightarrow \ln v + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 v - 1}{\sqrt{3}}) = - \ln x + c \Rightarrow \ln y + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - x}{\sqrt{3} x}) = c$

`⇒ 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.` $\ln y + \frac{2}{\sqrt{3}} \tan^{- 1} (\frac{2 y - x}{\sqrt{3} x}) = \frac{π}{\sqrt{3}}$

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Basic Concepts of Differential Equations

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