This checks, so we can proceed to solve for f: =M = ey hence f(x, y) = x.ey + ph

Question

This checks, so we can proceed to solve for f: =M = ey hence f(x, y) = x.ey + phi (y). But then f(x, y) = (x . ey + phi(y)) = x.ey + phi (y) And this last expression must equal N (x, y) = xey + 2y . It follows that Phi' (y) = 2y phi(y) = y2 + d. Altogether , then , we conclude that f(x, y) = x ey + y2 + d. We must not forget the final step. The solution of the differential equation is f(x, y) = c x ey + y2 = C This time we must content ourselves with the solution expressed implicitly , since it is not feasible to solve for y in terms of x. (y - x3)dx + (x + y3)dy = 0 (2y2 - 4ax + 5) dx = (4 - 2y + 4xy)dy (y + y cos xy) dx + (x + x cos xy) dy = 0 cos x cos2 y dx + 2 sin x sin y cos y dy =0 (sin x sin y - xey) dy = (ey + cos xcosy)dx -1 / y sin x / ydx + x / y2 sin x / y dy = 0 (1+y)dx + (1 -x)dy = 0 (2xy3 + y cos x) dx + (3x2y2 + sin x)dy =0 dx = y / 1 - x2y2dx + x / 1 - x2y2dy (2xy4 + sin y) dx + (4x2y3 + xcos y) dy = 0 ydx + xdy / 1 - x2y2 + xdx = 0

Explanation / Answer

Mdx + Ndy = 0

diff M wrt y : sinx sec^2 y

diff N wrt x : -sinx sec^2 y

not equal hence not EXACT

4) Mdx + Ndy = 0

4y

4y

EXACT

7) cosx siny - e^y  

-e^y -siny cosx   not exact

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