The general solution of the differential equa | vedclass

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Question

(D) Given the differential equation $\frac{dy}{dx} = 	an \left(\frac{y}{x}ight) + \frac{y}{x}$.Substitute $\frac{y}{x} = v$,which implies $y = vx$.

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

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(D) Given the differential equation $\frac{dy}{dx} = 	an \left(\frac{y}{x}ight) + \frac{y}{x}$.Substitute $\frac{y}{x} = v$,which implies $y = vx$. Differentiating with respect to $x$,we get $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the original equation: $v + x \frac{dv}{dx} = 	an(v) + v$.Simplifying,we get $x \frac{dv}{dx} = 	an(v)$.Separating the variables,we have $\frac{dv}{	an(v)} = \frac{dx}{x}$,which is $\cot(v) \, dv = \frac{dx}{x}$.Integrating both sides: $\int \cot(v) \, dv = \int \frac{dx}{x}$.This gives $\ln |\sin(v)| = \ln |x| + \ln |c|$.Taking the exponential of both sides,we get $\sin(v) = cx$.Substituting back $v = \frac{y}{x}$,the general solution is $\sin \left(\frac{y}{x}ight) = cx$.

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

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