2007-10-03 21:48:16 · 7 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
∫ dy / y = ∫ dx / x
log y = log x + log k
y = k x
2007-10-03 22:11:53 · answer #1 · answered by Como 7 · 2⤊ 2⤋
This is separable:
1/y dy/dx = 1/x (or 1/y dy = 1/x dx)
so integrating gives us ln |y| = ln |x| + c
=> y = ±e^c |x|
=> y = kx (we can get rid of the absolute value signs because, on either side of 0, we can absorb them into k as we have the ±).
2007-10-04 04:54:27 · answer #2 · answered by Scarlet Manuka 7 · 2⤊ 0⤋
sepation of variables take y yto left and x dx to right
dy/y = dx/x
integrate ln y = ln x + c
y = e^c x or Cx where C is another constant
2007-10-04 04:54:15 · answer #3 · answered by Mein Hoon Na 7 · 0⤊ 1⤋
dy/dx = y/x
dy/y = dx/x
integral dy/y = integral dx/x
ln|y| = ln |x| + c
y = (e^c)x of simpler let e^c = constant d;
y = dx
2007-10-04 04:55:36 · answer #4 · answered by william 2 · 0⤊ 1⤋
dy/dx = y/x
dy/y = dx/x
Taking indefinite integration:
ln|y| = ln|x|+C
y(x) = Cx, where C is an unknown constant.
2007-10-04 04:56:09 · answer #5 · answered by Anonymous · 0⤊ 1⤋
1. dy / dx = y / x;
2. (1 / y) (dy / dx) = (1 / x)
3. log (y) = log (x) + c;
2007-10-04 04:58:58 · answer #6 · answered by ravi_verma76 1 · 0⤊ 2⤋
dy/dx = y/x
dy/d = y
y = y
dy/dx = y/x
y/x = yx
x = x
I think there is no definite solution to this problem.
2007-10-04 04:58:54 · answer #7 · answered by Jun Agruda 7 · 2⤊ 1⤋