Question

Solve (x² + y²) dx + 2xy dy = 0

Answer 1

(x² + y²) dx + 2xy dy = 0

2xy dy = – (x² + y²) dx

`("d"y)/("d"x) = (-(x^2 + y^2))/(2xy)`  ........(1)

This is a homogeneous differential equation

Put y = vx and `("d"y)/("d"x) = "v" + x "dv"/("d"x)`

∴ (1) ⇒ `"v" + x "dv"/("d"x) = (-(x^2 + "v"^2x^2))/(2x("v"x))`

`"v" + x "dv"/("d"x) = (-x^2(1 + "v"^2))/(2"v"x^2)`

`"v" + x "dv"/("d"x) = (-(1 + "v"^2))/(2"v")`

`x "dv"/("d"x) = (-(1 + "v"^2)/(2"v") - "v"`

`x "dv"/("d"x) = (-1 - "v"^2 - 2"v"^2)/(2"v")`

= `(-3"v"^2 - 1)/(2"v")`

= `(-(3"v"^2+ 1))/(2"v")`

`((2"v"))/((3"v"^2 + 1)) "dv" = - 1/x  "d"x`

Integrating on both sides

`int ((2"v"))/((3"v"^2 + 1)) "dv" = - int 1/x  "d"x`

`1/3 int (6"v")/((3"v"^2 + 1)) "dv" = - int 1/x  "d"x`

`1/3 log (3"v"^2 + 1) = - log x + log "c"`

`log (3"v"^2 + 1)^(1/3) + log x = log "c"`

`log x (3"v"^2 + 1)^(1/3) = log "c"`

⇒ `x(3"v"^2 + 1)^(1/3)` = c

⇒ `x[(3y^2)/x^3 + 1]^(1/3)` = c