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(C) Given the homogeneous differential equation $\frac{dy}{dx} = -\frac{x^2+3y^2}{3x^2+y^2}$.Substitute $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{

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$\log_e|x+y| + \frac{2xy}{(x+y)^2} = 0$

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(C) Given the homogeneous differential equation $\frac{dy}{dx} = -\frac{x^2+3y^2}{3x^2+y^2}$.Substitute $y = vx$,then $\frac{dy}{dx} = v + x\frac{dv}{dx}$.$v + x\frac{dv}{dx} = -\frac{1+3v^2}{3+v^2}$.$x\frac{dv}{dx} = -\frac{1+3v^2}{3+v^2} - v = -\frac{1+3v^2+3v+v^3}{3+v^2} = -\frac{(v+1)^3}{3+v^2}$.Separating variables: $\frac{3+v^2}{(v+1)^3} dv = -\frac{dx}{x}$.Using partial fractions: $\frac{3+v^2}{(v+1)^3} = \frac{A}{v+1} + \frac{B}{(v+1)^2} + \frac{C}{(v+1)^3}$.$3+v^2 = A(v+1)^2 + B(v+1) + C = A(v^2+2v+1) + B(v+1) + C$.Comparing coefficients: $A=1$,$2A+B=0 \implies B=-2$,$A+B+C=3 \implies 1-2+C=3 \implies C=4$.So,$\int \left(\frac{1}{v+1} - \frac{2}{(v+1)^2} + \frac{4}{(v+1)^3}\right) dv = -\int \frac{dx}{x}$.$\ln|v+1| + \frac{2}{v+1} - \frac{2}{(v+1)^2} = -\ln|x| + C$.$\ln|v+1| + \ln|x| + \frac{2(v+1)-2}{(v+1)^2} = C$.$\ln|x(v+1)| + \frac{2v}{(v+1)^2} = C$.Since $v = \frac{y}{x}$,$\ln|x+y| + \frac{2(y/x)}{(1+y/x)^2} = C \implies \ln|x+y| + \frac{2xy}{(x+y)^2} = C$.Given $y(1)=0$,$\ln|1+0| + 0 = C \implies C=0$.Thus,the solution is $\ln|x+y| + \frac{2xy}{(x+y)^2} = 0$.

The solution of the differential equation $\frac{dy}{dx} = -\left(\frac{x^2+3y^2}{3x^2+y^2}\right)$,$y(1)=0$ is

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