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. - Mathematics and Statistics

Sum

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.

Solution

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 - 2e^("3x")

i.e. 3(x + 3y) = 2(1 - e^("3x")).

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