find the equation of a curve that passes through the point (2,4) and has slope
3y/(x-1) at any point on it.
2007-03-24 18:16:26 · 5 answers · asked by joe s 1 in Science & Mathematics ➔ Mathematics
The equation of the curve will be the integral of the equation of the slope, and the constant of integration can be determined from the given point. In this case, dy = 3y/(x-1) dx, or dy/y = 3/(x-1) dx, and from here the integration is straightforward.
2007-03-24 18:22:00 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The slope is given by 3y/(x-1). Thus dy/dx = 3y/(x-1)
Separating the variables leads to
dy/y = 3dx/(x-1)
Integrating both sides
ln y = 3ln(x-1) + lnC........Eqn 1
Applying the initial condition x=2 y=4 leads to C=4
Rearranging RHS in Eqn 1
ln y = ln(4(x-1)^3) leads to y=4(x-1)^3
2007-03-25 01:46:45 · answer #2 · answered by A S 4 · 0⤊ 0⤋
y' = 3y/(x-1)
ln|y| = 3ln|x-1|+c*
y = c(x-1)^3
4 = c
The equation is,
y = 4(x-1)^3
Check:
y' = 12(x-1)^2 = 3y/(x-1)
It works.
2007-03-25 01:24:17 · answer #3 · answered by sahsjing 7 · 0⤊ 0⤋
(y - 4) = 3y(x - 2)/(x - 1)
(y - 4) (x - 1) = 3y (x - 2)
xy - y - 4x + 4 = 3xy - 6y
4 - 4x = 2xy - 5y
4 - 4x = y(2x - 5)
y = 4(1 - x)/(2x - 5)
2007-03-25 01:43:40 · answer #4 · answered by Helmut 7 · 0⤊ 0⤋
i know the answer to this one! i swear!! only the "top contributor" above me did it faster.. phooey! at this rate ill never be able to be a top contributor :(
2007-03-25 01:24:23 · answer #5 · answered by greenprincess 5 · 0⤊ 0⤋