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प्रश्न
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)`.
उत्तर
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.
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