The general solution of the differential equa | vedclass

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(B) Given equation: $x \cos \left(\frac{y}{x}\right)(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$Rearranging terms: $x y \cos \left(\frac{y}{x}\right

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$\cos \left(\frac{y}{x}\right)=\frac{C}{x y}$

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(B) Given equation: $x \cos \left(\frac{y}{x}ight)(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$Rearranging terms: $x y \cos \left(\frac{y}{x}ight) d x + x^2 \cos \left(\frac{y}{x}ight) d y = x y \sin \left(\frac{y}{x}ight) d y - y^2 \sin \left(\frac{y}{x}ight) d x$Grouping $dx$ and $dy$ terms: $[x y \cos \left(\frac{y}{x}ight) + y^2 \sin \left(\frac{y}{x}ight)] d x = [x y \sin \left(\frac{y}{x}ight) - x^2 \cos \left(\frac{y}{x}ight)] d y$$\frac{d y}{d x} = \frac{x y \cos \left(\frac{y}{x}ight) + y^2 \sin \left(\frac{y}{x}ight)}{x y \sin \left(\frac{y}{x}ight) - x^2 \cos \left(\frac{y}{x}ight)}$Divide numerator and denominator by $x^2$: $\frac{d y}{d x} = \frac{\frac{y}{x} \cos \left(\frac{y}{x}ight) + (\frac{y}{x})^2 \sin \left(\frac{y}{x}ight)}{\frac{y}{x} \sin \left(\frac{y}{x}ight) - \cos \left(\frac{y}{x}ight)}$Substitute $y=vx$,so $\frac{d y}{d x} = v + x \frac{d v}{d x}$:$v + x \frac{d v}{d x} = \frac{v \cos v + v^2 \sin v}{v \sin v - \cos v}$$x \frac{d v}{d x} = \frac{v \cos v + v^2 \sin v - v(v \sin v - \cos v)}{v \sin v - \cos v} = \frac{2v \cos v}{v \sin v - \cos v}$Separating variables: $\frac{v \sin v - \cos v}{v \cos v} d v = 2 \frac{d x}{x}$$\int (	an v - \frac{1}{v}) d v = 2 \int \frac{d x}{x}$$\log |\sec v| - \log |v| = 2 \log |x| + \log |C_1|$$\log |\frac{\sec v}{v}| = \log |C_1 x^2| \Rightarrow \frac{\sec v}{v} = C_1 x^2$Substituting $v = \frac{y}{x}$: $\frac{\sec(y/x)}{y/x} = C_1 x^2 \Rightarrow \frac{\sec(y/x)}{y} = C_1 x \Rightarrow \sec(y/x) = C_1 x y$Thus,$\frac{1}{\cos(y/x)} = C_1 x y \Rightarrow \cos(y/x) = \frac{1}{C_1 x y} = \frac{C}{x y}$.

The general solution of the differential equation $x \cos \frac{y}{x}(y d x+x d y)=y \sin \frac{y}{x}(x d y-y d x)$ is

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