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Question
Find the equation of a curve passing through origin if the slope of the tangent to the curve at any point (x, y) is equal to the square of the difference of the abcissa and ordinate of the point.
Solution
Here, slope of the tangent of the curve = `("d"y)/("d"x)` and the difference between the abscissa and ordinate = x – y.
∴ As per the condition, `("d"y)/("d"x) = (x - y)^2`
Put x – y = v
`1 - ("d"y)/("d"x) = "dv"/("d"x)`
∴ `("d"y)/("d"x) = 1 - "dv"/"dx"`
∴ The equation becomes `1 - "dv"/"dx" = "v"^2`
⇒ `"dv"/"dx" = 1 - "v"^2`
⇒ `"dv"/(1 - "v"^2)` = dx
Integrating both sides, we get
`int "dv"/(1 - "v"^2) = int "d"x`
⇒ `1/2 log |(1 + "v")/(1 - "v")|` = x + c
⇒ `1/2 log|(1 + x - y)/(1 - x + y)|` = x + c ......(1)
Since, the curve is passing through (0, 0)
Then `1/2 log|(1 + 0 - 0)/(1 - 0 + 0)|` = 0 + c
⇒ c = 0
∴ On putting c = 0 in equation (1) we get
`1/2 log |(1 + x - y)/(1 - x + y)|` = x
⇒ `log|(1 + x - y)/(1 - x + y)|` = 2x
∴ `(1 + x - y)/(1 - x + y)|` = e2x
⇒ (1 + x – y) = e2x (1 – x + y)
Hence, the required equation is (1 + x – y) = e2x (1 – x + y).
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