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Question
Solve the following differential equation:
(x2 + y2) dy = xy dx. It is given that y (1) = y(x0) = e. Find the value of x0
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Solution
The given differential equation is of the form
`("d"y)/("d"x) = (xy)/(x^2 + y^2)` ........(1)
This is a homogeneous differential equation.
Putting y = vx
`("d"y)/("d"x) = "v"(1) + x "dv"/("d"x)`
`("d"y)/("d"x) = "v" + x "dv"/("d"x)`
Now equation (1) becomes,
`"v" + x "dv"/("d"x) = (x("v"x))/(x^2 + ("v"x)^2) = (x^2"v")/(x^2[1 + "v"^2])`
`x "dv"/("d"x) = "v"/(1 + "v"^2) - "v"`
`x "dv"/("d"x) = ("v" - "v" - "v"^3)/(1 + "v"^2)`
`x "dv"/("d"x) = (-"v"^3)/(1 + "v"^2)`
`((1 + "v"^2)"dv")/"v"^3 = (-"d"x)/x`
`1/"v"^3 "dv" + "v"^2/"v"^3 "dv" = (- "d"x)/x`
`int "v"^-3 "dv" + int 1/"v" "dv" = - int ("d"x)/x`
`"v"^-2/(-2) + log "v" = - logx + log"c"`
`1/(2"v"^2) - log"v" = log x - log "c"`
`1/(2"v"^2) = logx - log"c" + log"v"`
`1/(2"v"^2) = log ("v"x)/"c"`
∵ y = vx
⇒ v = `y/x`
∴ `x^2/(2y^2) = log y/"c"`
`"e"^(x^2/(2y^2)) = y/"c"`
⇒ y = `"Ce"^(x^2/(2y^2))` ........(2)
Given y(1) = 1
i.e., when x = 1, y = 1
(2) ⇒ 1 = `"Ce"^(1/2)`
C = `1/sqrt("e")`
Now (2 becomes y = `1/sqrt("e") "e"^(x^3/(2y^2))`
Also given y(x0) = e
i.e., when x = x0, y = e
∴ e = `1/sqrt("e") "e"^((x_0^2)/(2"e"^2))`
`"e"sqrt("e") = "e"^((x_0^2)/(2"e"^2))`
`log "e"sqrt("e") = (x_0^2)/(2"e"^2)`
`log "e"^(3/2) = (x_0^2)/(2"e"^2)`
`3/2 log "e" = (x_0^2)/(2"e"^2)`
`x_0^2 = 3"e"^2`
x0 = `+- sqrt(3)*"e"`
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