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(A) Given the differential equation $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$.Taking the reciprocal,we get $\frac{dy}{dx} = \frac{1 + x e^y \cos

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$2x + e^y(c + \sin(x^2)) = 0$

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(A) Given the differential equation $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$.Taking the reciprocal,we get $\frac{dy}{dx} = \frac{1 + x e^y \cos(x^2)}{x} = \frac{1}{x} + e^y \cos(x^2)$.Rearranging the terms,we have $\frac{dy}{dx} - e^y \cos(x^2) = \frac{1}{x}$.This is not a standard linear form. Let's re-examine the original equation: $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$.$\frac{dy}{dx} = \frac{1 + x e^y \cos(x^2)}{x} = \frac{1}{x} + e^y \cos(x^2)$.Divide by $e^y$: $e^{-y} \frac{dy}{dx} - \cos(x^2) = \frac{1}{x} e^{-y}$.$e^{-y} \frac{dy}{dx} - \frac{e^{-y}}{x} = \cos(x^2)$.Let $v = -e^{-y}$,then $\frac{dv}{dx} = e^{-y} \frac{dy}{dx}$.Substituting this into the equation: $\frac{dv}{dx} + \frac{v}{x} = \cos(x^2)$.This is a linear differential equation in $v$ with integrating factor $I.F. = e^{\int \frac{1}{x} dx} = e^{\ln x} = x$.The solution is $v \cdot x = \int x \cos(x^2) dx$.Let $u = x^2$,then $du = 2x dx$,so $x dx = \frac{1}{2} du$.$v \cdot x = \int \frac{1}{2} \cos(u) du = \frac{1}{2} \sin(u) + c = \frac{1}{2} \sin(x^2) + c$.Substituting $v = -e^{-y}$: $-x e^{-y} = \frac{1}{2} \sin(x^2) + c$.$-2x = e^y (\sin(x^2) + 2c)$.Rearranging gives $2x + e^y(C + \sin(x^2)) = 0$.

The solution of the differential equation $\frac{dx}{dy} = \frac{x}{1 + x e^y \cos(x^2)}$ is (where $c$ is the constant of integration):

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