Any help to find the initial value problem of:
dy/dx= -(2ycosx)^2
2007-01-04 18:21:11 · 3 answers · asked by Yorkie1961 2 in Science & Mathematics ➔ Mathematics
A little rusty, but looks like a clean separation of variables problem, so just move the y terms to one side, the x terms to the other, and integrate both sides. I assume you have additional information to solve for the integration constant which appears.
1/y^2 * dy = -4(cosx)^2 * dx
- 1/y = -4 * (x/2 + sin2x/4) + C
or 1/y = 2x + sin2x + C
So plug in the IVP values of y and x to solve for C.
2007-01-04 19:17:06 · answer #1 · answered by SAN 5 · 0⤊ 0⤋
Check out the back cover of your calculus textbook.
2007-01-04 18:28:35 · answer #2 · answered by strong_perkasa 2 · 0⤊ 1⤋
You need help finding the porblem?
Or do you want help finding the answer?
2007-01-04 18:24:33 · answer #3 · answered by ? 6 · 0⤊ 0⤋