Question

Solve the following homogeneous differential equation:

`x ("d"y)/("d"x) - y = sqrt(x^2 + y^2)`

Answer 1

`x ("d"y)/("d"x) - y = sqrt(x^2 + y^2)`

`x ("d"y)/("d"x) = sqrt(x^2 + y^2) + y`

`("d"y)/("d"x) = sqrt(x^2 + y^2 + y)/x`  ........(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) = (sqrt(x^2 + ("v"x)^2) + "v"x)/x`

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

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

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

⇒ `x "dv"/("d"x) = sqrt(1 + "v"^2)`

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

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

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

⇒  `"v" + sqrt(1 + "v"^2)` = xc

`y/x + sqrt(1 + y^2/x^2)` = xc

`y/x + sqrt((x^2 + y^2)/x^2)` = xc

⇒ `y/x + sqrt(x^2 + y^2)/x^2` = xc

`1/x [y + sqrt(y^2 + y^2)]` = xc

⇒ `y + sqrt(x^2 + y^2)` = x²c