Solve the differential equation (y-xy')/(x+yy') = 2

Solve the differential equation $\frac{y-xy'}{x+yy'} = 2$.

Explanation of Method

Cross-multiplying yields a homogeneous first-order ODE, which is solved with the substitution $y=vx$, reducing it to a separable equation whose integral produces an implicit solution in polar-like form $\arctan(y/x)+\ln(x^2+y^2)=C$.

Rewrite as an explicit ODE

Cross-multiplying: $y-xy' = 2(x+yy') = 2x+2yy'$, so $y-2x = y'(x+2y)$, giving

\[y' = \frac{y-2x}{x+2y}.\]

Both numerator and denominator are homogeneous of degree 1, so the equation is homogeneous.

Substitute $y=vx$

With $y=vx$, $y'=v+xv'$, and

\[v+xv' = \frac{vx-2x}{x+2vx} = \frac{v-2}{1+2v}.\]

Hence

\[xv' = \frac{v-2}{1+2v} - v = \frac{-2(1+v^2)}{1+2v}.\]

Separate and integrate

\[\frac{1+2v}{1+v^2}\,dv = -\frac{2\,dx}{x}.\]

Splitting the left side, $\displaystyle\int\frac{dv}{1+v^2}=\arctan v$ and $\displaystyle\int\frac{2v\,dv}{1+v^2}=\ln(1+v^2)$, so

\[\arctan v + \ln(1+v^2) = -2\ln|x| + C_1.\]

Back-substitute $v=y/x$

Absorbing $-2\ln|x|$ and $\ln x^2$ together: $\arctan\frac{y}{x} + \ln\bigl[x^2(1+y^2/x^2)\bigr] = C_1$, i.e.

\[\arctan\frac{y}{x} + \ln(x^2+y^2) = C.\]

Verify by implicit differentiation

Differentiating $\arctan(y/x)+\ln(x^2+y^2)=C$ implicitly (verified with sympy) gives $\dfrac{xy'-y}{x^2+y^2} + \dfrac{2x+2yy'}{x^2+y^2}=0$, which simplifies to $y'(x+2y) = y-2x$, i.e. $y'=\dfrac{y-2x}{x+2y}$ — exactly the ODE from Step 1. This also matches solving the original equation $\frac{y-xy'}{x+yy'}=2$ directly for $y'$.

Result

Final answer:

\[\boxed{\arctan\frac{y}{x} + \ln(x^2+y^2) = C}\]

Equivalently, $x^2+y^2 = Ke^{-\arctan(y/x)}$ with $K=e^C>0$. Verified by implicit differentiation reproducing $y' = \dfrac{y-2x}{x+2y}$ exactly. The source's derivation and final answer are correct as given.