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प्रश्न
If the slope of the tangent to the curve at each of its point is equal to the sum of abscissa and the product of the abscissa and ordinate of the point. Also, the curve passes through the point (0, 1). Find the equation of the curve.
उत्तर
Let A (x, y) be any point on the curve y = f(x).
Then the slope of the tangent to the curve at point A is `"dy"/"dx"`.
According to the given condition
`"dy"/"dx"` = x + xy
∴ `"dy"/"dx"` - xy = x .....(1)
This is the linear differential equation of the form
`"dy"/"dx" + "P" "y" = "Q"`, where P = - x and Q = x
∴ I.F. = `"e"^(int "P dx") = "e"^(int -"x" "dx") = "e"^(-"x"^2/2)`
∴ the solution of (1) is given by
`"y"*("I.F.") = int "Q" * ("I.F.") "dx" + "c"`
∴ `"y" * "e"^(-"x"^2/2) = = int "x" * "e"^(-"x"^2/2) "dx" + "c"`
∴ `"e"^(-"x"^2/2) * "y" = int "x" * "e"^(-"x"^2/2) "dx" + "c"`
∴ Put `- "x"^2/2 = "t"`
∴ - x dx = dt
∴ x dx = - dt
∴ `"e"^(-"x"^2/2) * "y" = int "e"^"t" * (- "dt") + "c"`
∴ `"e"^(-"x"^2/2) * "y" = - int "e"^"t" * "dt" + "c"`
∴ `"e"^(-"x"^2/2) * "y" = - "e"^"t" + "c"`
∴ `"e"^(-"x"^2/2) * "y" = - "e"^(-"x"^2/2) + "c"`
∴ y = - 1 + `"ce"^("x"^2/2)`
∴ 1 + y = `"ce"^("x"^2/2)` .....(2)
This is the general equation of the curve.
But the required curve is passing through the point (0, 1).
∴ by putting x = 0 and y = 1 in (2), we get
1 + 1 = c
∴ c = 2
∴ from (2), the equation of the required curve is
1 + y = `2"e"^("x"^2/2)`.
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