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प्रश्न
Find the equation of the curve which passes through the origin and has the slope x + 3y - 1 at any point (x, y) on it.
उत्तर
Let A (x, y) be the point on the curve y = f(x).
Then slope of the tangent to the curve at point A is `"dy"/"dx"`.
According to the given condition,
`"dy"/"dx" = "x" + 3"y" - 1`
∴ `"dy"/"dx" - 3"y" = "x - 1"` ....(1)
This is the linear differential equation of the form
`"dy"/"dx" + "Py" = "Q"`, where P = - 3 and Q = x - 1
∴ I.F. = `"e"^(int "P dx") = "e"^(int - 3"dx") = "e"^(- 3"x")`
∴ the solution of (1) is given by
`"y" * ("I.F.") = int "Q" * ("I.F.") "dx" + "c"`
∴ `"y" * "e"^(- 3"x") = int ("x - 1") * "e"^(-3"x") "dx" + "c"`
∴ `"e"^(- 3"x") * "y" = ("x - 1") int "e"^(- 3"x") - int ["d"/"dx" ("x - 1") * int "e"^(- 3"x") "dx"] "dx" + "c"_1`
∴ `"e"^(- 3"x") * "y" = ("x - 1") * "e"^(- 3"x")/-3 - int 1 * "e"^(- 3"x") * "y"/-3 "dx" + "c"_1`
∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") * "e"^(- 3"x") + 1/3 int "e"^(- 3"x") "dx" + "c"_1`
∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1") "e"^(- 3"x") + 1/3 * "e"^(- 3"x")/-3 + "c"_1`
∴ `"e"^(- 3"x") * "y" = - 1/3 ("x - 1")"e"^(- 3"x") - 1/9 "e"^(- 3"x") + "c"_1`
∴ 9y = - 3(x - 1) - 1 + 9`"c"_1 * e^("3x")`
∴ 9y + 3(x - 1) + 1 = `9"c"_1 * e^("3x")`
∴ 9y + 3x - 3 + 1 = `9"c"_1 * e^("3x")`
∴ 3(x + 3y) = 2 + `9"c"_1 * e^("3x")`
∴ 3(x + 3y) = 2 + `"c" * e^("3x")` where c = 9c1 ....(2)
This is the general equation of the curve.
But the required curve is passing through the origin (0, 0).
∴ by putting x = 0 and y= 0 in (2), we get
0 = 2 + c
∴ c = - 2
∴ from (2), the equation of the required curve is
3(x + 3y) = 2 - 2`e^("3x")`
i.e. 3(x + 3y) = 2(1 - `e^("3x")`).
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