Question

Find the equation of a curve passing through (2, 1) if the slope of the tangent to the curve at any point (x, y) is `(x^2 + y^2)/(2xy)`.

Answer 1

Given that the slope of tangent to a curve at (x, y) is `("d"y)/("d"x) = (x^2 + y^2)/(2xy)`

It is a homogeneous differential equation

So, put y = vx

⇒ `("d"y)/("d"x) = "v" + x * "dv"/"dx"`

`"v" + x * "dv"/"dx" = (x^2 + "v"^2x^2)/(2x * "v"x)`

⇒ `"v" + x * "dv"/"dx" = (1 + "v"^2)/(2"v")`

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

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

⇒ `x * "dv"/"dx" = (1 - "v"^2)/(2"v")`

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

Integrating both sides, we get

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

⇒ `-log|1 - "v"^2| = log x + log "c"`

⇒ `-log|1 - y^2/x^2| = logx + log"c"`

⇒ `-log|(x^2 - y^2)/x| = logx + log"c"`

⇒ `log|x^2/(x^2 - y^2)| = log|x"c"|`

⇒ `x^2/(x^2 - y^2)` = xc

Since, the curve is passing through the point (2, 1)

∴ `(2)^2/((2)^2 - (1)^2` = 2c

⇒ `4/3` = 2c

⇒ c = `2/3`

Hence, the required equation is `x^2/(x^2 - y^2) = 2/3 x`

⇒ 2(x² – y²) = 3x.