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प्रश्न
The curve passes through the point (0, 2). The sum of the coordinates of any point on the curve exceeds the slope of the tangent to the curve at any point by 5. Find the equation of the curve.
उत्तर
Let A(x, y) be any point on the curve.
Then slope of the tangent to the curve at point A is `"dy"/"dx"`.
According to the given condition
x + y = `"dy"/"dx" + 5`
∴ `"dy"/"dx" - "y" = "x - 5"` ...(1)
This is the linear differential equation of the form
`"dy"/"dx" + "P" * "y" = "Q"`, where P = - 1 and Q = x - 5
∴ I.F. = `"e"^(int "P dx") = "e"^(int -1 "dx") = "e"^-"x"`
∴ the solution of (1) is given by
`"y"*("I.F.") = int "Q" * ("I.F.") "dx" + "c"`
∴ `"y" * "e"^-"x" = int ("x - 5")"e"^-"x" "dx" + "c"`
∴ `"e"^-"x" * "y" = ("x - 5") int "e"^-"x" "dx" - int ["d"/"dx" ("x - 5") int "e"^-"x" "dx"] "dx" + "c"`
∴ `"e"^-"x" * "y" = ("x - 5") * "e"^-"x"/-1 - int 1 * "e"^-"x"/-1 "dx" + "c"`
∴ `"e"^-"x" * "y" = - ("x - 5") * "e"^-"x" + int "e"^-"x" "dx" + "c"`
∴ `"e"^-"x" * "y" = - ("x - 5")"e"^-"x" + "e"^-"x"/-1 + "c"`
∴ y = - (x - 5) - 1 + cex
∴ y = - x + 5 - 1 + cex
∴ y = 4 - x + cex ....(2)
This is the general equation of the curve.
But the required curve is passing through the point (0, 2).
∴ by putting x = 0, y = 2 in (2), we get
2 = 4 - 0 + c
∴ c = - 2
∴ from (2), the equation of the required curve is
y = 4 - x - 2ex
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