The solution of frac{dy}{dx} = frac{y}{x} + t | vedclass

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + 	an \frac{y}{x}$.This is a homogeneous differential equation.Let $v = \frac{y}{x}

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$x = c \sin(y/x)$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{y}{x} + 	an \frac{y}{x}$.This is a homogeneous differential equation.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 equation: $v + x \frac{dv}{dx} = v + 	an v$.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: $\int \cot v \, dv = \int \frac{dx}{x}$.This gives $\ln |\sin v| = \ln |x| + \ln |c|$.Using logarithmic properties: $\ln |\sin v| = \ln |cx|$.Taking the exponential of both sides: $\sin v = cx$.Substituting back $v = \frac{y}{x}$: $\sin(\frac{y}{x}) = cx$, or $x = c \sin(\frac{y}{x})$.

The solution of $\frac{dy}{dx} = \frac{y}{x} + \tan \frac{y}{x}$ is

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