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

Answer 1

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"`