Advertisements
Advertisements
प्रश्न
Find the curve whose gradient at any point P(x, y) on it is `(x - "a")/(y - "b")` and which passes through the origin
उत्तर
The gradient of the curve at P(x, y)
`("d"y)/("d"x) = (x - "a")/(y - "b")`
(y – b) dy = (x – a) dx
Integrating on both sides
`int (y - "b") "d"y = int (x - "a") "d"x`
⇒ `(y -- "b")^2/2 = (x - )^2/2 + "c"`
Multiply each term by 2
∴ (y – b)2 = (x – a)2 + 2c .........(1)
Since the curve passes through the origin (0, 0)
Equation (1)
(0 – b)2 = (0 – a)2 + 2c
b2 = a2 + 2c
b2 – a2 = 2c .......(2)
Substitute equation (2) and (1)
(y – b)2 = (x – a)2 + b2 – a2
APPEARS IN
संबंधित प्रश्न
The velocity v, of a parachute falling vertically satisfies the equation `"v" (dv)/(dx) = "g"(1 - v^2/k^2)` where g and k are constants. If v and are both initially zero, find v in terms of x
Solve the following differential equation:
`(1 + 3"e"^(y/x))"d"y + 3"e"^(y/x)(1 - y/x)"d"x` = 0, given that y = 0 when x = 1
Choose the correct alternative:
The solution of `("d"y)/("d"x) = 2^(y - x)` is
Solve: ydx – xdy = 0 dy
Solve: `("d"y)/("d"x) = y sin 2x`
Solve the following:
`("d"y)/("d"x) + (3x^2)/(1 + x^3) y = (1 + x^2)/(1 + x^3)`
Solve the following:
`("d"y)/("d"x) + y/x = x'e"^x`
Choose the correct alternative:
The differential equation of x2 + y2 = a2
Solve (D2 – 3D + 2)y = e4x given y = 0 when x = 0 and x = 1
Solve x2ydx – (x3 + y3) dy = 0
