x^2 d^2x/dy^2 + 2xdx/dy=0
2006-12-03 00:47:30 · 5 answers · asked by Glendon 1 in Science & Mathematics ➔ Mathematics
11 and 24, Bloody hell im a genus...
2006-12-03 00:52:53 · answer #1 · answered by Loving and loved 2 · 0⤊ 1⤋
15
2006-12-03 00:56:47 · answer #2 · answered by Anonymous · 0⤊ 0⤋
Your D.E. is
x^2 y'' + 2x y' = 0
If you assume that y=x^n then
x^2 n(n-1) x^(n-2) + 2x n x^(n-1) = 0 or
n(n-1) x^n + 2n x^n = 0 or
x^n ( n(n-1) + 2n) = 0
You want to find values of n so that this is true for all x which implies that
n(n-1) + 2n = 0 or
n [ (n-1)+2] = 0
n (n+1) = 0
So n=0 or n=-1
Solutions of the ODE are of the form
y = C + Dx^(-1)
where C and D are constants.
2006-12-03 01:05:05 · answer #3 · answered by hij 2 · 0⤊ 0⤋
divide throughout by x^2,
d^2 y / dx^2 + (2/x) dy/dx =0
Now get the complementary function and you're done
2006-12-03 01:08:35 · answer #4 · answered by yasiru89 6 · 0⤊ 0⤋
right it quite is a sprint: Use a substitution the place y = xu and dy = xdu + udx 2x²dx + 2y²dx = xydy 2x²dx + 2x²u²dx = x²u(xdu + udx) 2x²dx + 2x²u²dx = x²u²dx + x³udu -x³udu = x²u²dx - 2x²dx - 2x²u²dx -x³udu = -x²u²dx - 2x²dx -x³udu = -x²(u² + 2)dx udu / (u² + 2) = dx / x ?(u / (u² + 2))du = ?dx / x i'm going to provide help to end it from right here.
2016-10-17 15:40:36 · answer #5 · answered by ? 4 · 0⤊ 0⤋