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ysin2xdx-(1+y²+cos²x)dy=0
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The equation ysin2xdx + (y^2 - 1 - (cosx)^2)dy =0 is exact because
(ysin2x)_y = sin2x = (y^2 -1 - (cosx)^2)_x . Then the equation is the total
differential dF(x,y) =0 with solution F(x,y) = C obtained from the system
F_x = ysin2x
F_y = y^2 -1 - (cosx)^2. Integrating the first Eq. leads to
F(x,y) = -(ycos2x)/2 + G(y) = y( 1/2 - (cosx)^2) + G(y).The function G(y)
is obtained from the second Eq. which gives G’(y) = y^2 - 3/2 . Then
G(y) = (1/3)y^3 - 3y/2 + C and the solution is
F(x,y) = (1/3)y^3 - (y/2)cos2x - 3y/2 =C.
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