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(A) Given the differential equation: $\frac{dy}{dx} = \frac{2x+3y}{3x-2y}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{

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$\frac{1}{3} \log(x^2+y^2)$

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(A) Given the differential equation: $\frac{dy}{dx} = \frac{2x+3y}{3x-2y}$.This is a homogeneous differential equation. Let $y = vx$, then $\frac{dy}{dx} = v + x \frac{dv}{dx}$.Substituting these into the equation:$v + x \frac{dv}{dx} = \frac{2x + 3vx}{3x - 2vx} = \frac{2+3v}{3-2v}$.$x \frac{dv}{dx} = \frac{2+3v}{3-2v} - v = \frac{2+3v - 3v + 2v^2}{3-2v} = \frac{2v^2+2}{3-2v}$.Separating variables:$\frac{3-2v}{2(v^2+1)} dv = \frac{dx}{x}$.Integrating both sides:$\frac{3}{2} \int \frac{dv}{v^2+1} - \int \frac{v}{v^2+1} dv = \int \frac{dx}{x}$.$\frac{3}{2} 	an^{-1}(v) - \frac{1}{2} \ln(v^2+1) = \ln|x| + C$.$\frac{3}{2} 	an^{-1}(v) = \ln|x| + \frac{1}{2} \ln(v^2+1) + C = \ln|x \sqrt{v^2+1}| + C = \ln \sqrt{x^2+y^2} + C$.$	an^{-1}(\frac{y}{x}) = \frac{1}{3} \ln(x^2+y^2) + C'$.$\frac{y}{x} = 	an(\frac{1}{3} \ln(x^2+y^2) + C')$.Comparing with $y = x 	an(f(x)) + c$, we get $f(x) = \frac{1}{3} \log(x^2+y^2)$.

If the solution of the differential equation $\frac{dy}{dx} = \frac{2x+3y}{3x-2y}$ is $y = x \tan(f(x)) + c$, then $f(x) =$

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