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(B) Given the differential equation: $\frac{dy}{dx} = 	an \left(\frac{y}{x}ight) + \frac{y}{x}$.Let $v = \frac{y}{x}$,so $y = vx$.Differentiating w

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$\sin \left(\frac{y}{x}\right) = cx$

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(B) Given the differential equation: $\frac{dy}{dx} = 	an \left(\frac{y}{x}ight) + \frac{y}{x}$.Let $v = \frac{y}{x}$,so $y = vx$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the original equation:$v + x \frac{dv}{dx} = 	an v + v$.Subtracting $v$ from both sides:$x \frac{dv}{dx} = 	an v$.Separating the variables:$\frac{dv}{	an v} = \frac{dx}{x}$,which is $\cot v \, dv = \frac{dx}{x}$.Integrating both sides:$\int \cot v \, dv = \int \frac{1}{x} \, dx$.$\log |\sin v| = \log |x| + \log |c|$.$\log |\sin v| = \log |cx|$.Taking the exponential of both sides:$\sin v = cx$.Substituting $v = \frac{y}{x}$ back:$\sin \left(\frac{y}{x}ight) = cx$.

The solution of the differential equation $\frac{dy}{dx} = \tan \left(\frac{y}{x}\right) + \frac{y}{x}$ is

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