Question

Solve the following homogeneous differential equation:

`(x - y) ("d"y)/("d"x) = x + 3y`

Answer 1

`("d"y)/("d"x) = (x + 3y)/((x - y))`  ........(1)

It is a homogeneous differential equation same degree in x and y

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

Equation (1)

⇒ `"v" + x "dv"/("d"x) = ((x + 3 "v"x))/((x - "v"x))`

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

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

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

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

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

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

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

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

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

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

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

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

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

Integrating on both sides

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

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

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

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

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

`(-2)/((x + y)) = log "c" ((x + y)/x)`

`((-2x)/(x + y)) = log "c" [x + y]`

⇒ `"e"^(- (2x)/(x + y)) = "c"(x + y)`

`1/"c"  "e"^(-(2x)/(x + y)) = (x + y)` .....`{"Let" 1/"c"  "be"  "k"}`

⇒ `"ke"^(-(2x)/(x + y)) = (x + y)`

⇒ x + y = `"ke"^(-(2x)/(x + y))`