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(B) Given differential equation is $y dx = (x + y^3 \cos y) dy$. Dividing by $y dy$,we get $\frac{dx}{dy} = \frac{x}{y} + y^2 \cos y$. Rearranging the

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$x = y^2 \sin y + 2y \cos^2 \frac{y}{2}$

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(B) Given differential equation is $y dx = (x + y^3 \cos y) dy$. Dividing by $y dy$,we get $\frac{dx}{dy} = \frac{x}{y} + y^2 \cos y$. Rearranging the terms,we get $\frac{dx}{dy} - \frac{1}{y} x = y^2 \cos y$. This is a linear differential equation of the form $\frac{dx}{dy} + P(y)x = Q(y)$,where $P(y) = -\frac{1}{y}$ and $Q(y) = y^2 \cos y$. The integrating factor $IF = e^{\int P(y) dy} = e^{\int -\frac{1}{y} dy} = e^{-\ln y} = \frac{1}{y}$. The solution is $x \cdot IF = \int Q(y) \cdot IF dy + C$. $x \cdot \frac{1}{y} = \int (y^2 \cos y) \cdot \frac{1}{y} dy + C$. $\frac{x}{y} = \int y \cos y dy + C$. Using integration by parts,$\int y \cos y dy = y \sin y - \int \sin y dy = y \sin y + \cos y$. So,$\frac{x}{y} = y \sin y + \cos y + C$. $x = y^2 \sin y + y \cos y + Cy$. The curve passes through $(0, \pi)$,so $0 = \pi^2 \sin \pi + \pi \cos \pi + C\pi$. $0 = 0 - \pi + C\pi \implies C\pi = \pi \implies C = 1$. Thus,$x = y^2 \sin y + y \cos y + y = y^2 \sin y + y(1 + \cos y) = y^2 \sin y + y(2 \cos^2 \frac{y}{2})$. Therefore,$x = y^2 \sin y + 2y \cos^2 \frac{y}{2}$.

The equation of the curve passing through the point $(0, \pi)$ and satisfying the differential equation $y dx = (x + y^3 \cos y) dy$ is

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