Find the general solution of: 2xsin3ydx + (3x^2cos3y+2y)dy = 0 Solution => 2 x s

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=> 2 x sin(3 y(x))+(3 x^2 cos(3 y(x))+2 y(x)) ( dy(x))/( dx) = 0 Let P(x, y) = 2 sin(3 y) x and Q(x, y) = 2 y+3 cos(3 y) x^2. This is an exact equation, because (dP(x, y))/(dy) = 6 cos(3 y) x = (dQ(x, y))/(dx). f(x, y) = integral 2 sin(3 y) x dx = sin(3 y) x^2+g(y) where g(y) is an arbitrary function of y. Differentiating- (df(x, y))/(dy) = (d)/(dy)(sin(3 y) x^2+g(y)) = 3 cos(3 y) x^2+( dg(y))/( dy) Substitute into (df(x, y))/(dy) = Q(x, y): 3 cos(3 y) x^2+( dg(y))/( dy) = 2 y+3 cos(3 y) x^2 ( dg(y))/( dy) = 2 y Integrate ( dg(y))/( dy) with respect to y: g(y) = integral 2 y dy = y^2 Substitute g(y) into f(x, y): f(x, y) = y^2+sin(3 y) x^2 The solution is f(x, y) = c: y^2+sin(3 y) x^2 = c

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