Given the differential equation y'+(x)/(y)=1 . 1^° By…

Given the differential equation \(y'+\dfrac{x}{y}=1\). \(1^\circ\) By substituting \(y=vx\), show that the solution is defined by \[ \frac{\sqrt3}{2}\ln(x^2-xy+y^2)+\arctan\frac{2y-x}{x\sqrt3}=c_1. \] \(2^\circ\) By substituting \(x=vy\), show that the solution is defined by \[ \frac{\sqrt3}{2}\ln(x^2-xy+y^2)-\arctan\frac{2x-y}{y\sqrt3}=c_2. \] \(3^\circ\) Show that these solutions are equivalent and find the relationship between \(c_1\) and \(c_2\).

1. Confirm Part \(1^\circ\)'s relation solves \(y'+x/y=1\)

With \(F_1(x,y)=\frac{\sqrt3}{2}\ln(x^2-xy+y^2)+\arctan\frac{2y-x}{x\sqrt3}\), Sympy computes \(\dfrac{d}{dx}F_1(x,y(x))\), sets it to \(0\), and solves for \(y'(x)\); the result simplifies exactly to \(1-x/y\). So the substitution \(y=vx\) (source's steps 1–5) does lead to a correct implicit solution, even though the source's own hand-integration in step 3 visibly stumbles ("Actually, let's use…").

2. Confirm Part \(2^\circ\)'s relation solves \(y'+x/y=1\)

The same implicit-differentiation check applied to \(F_2(x,y)=\frac{\sqrt3}{2}\ln(x^2-xy+y^2)-\arctan\frac{2x-y}{y\sqrt3}\) again returns \(y'=1-x/y\) exactly. So both families are genuine first integrals of the same ODE — the source's step 7 ("following similar steps…") is confirmed correct even though it was never actually carried out there.

3. Reduce \(c_1-c_2\) to a function of \(v=y/x\) alone, and show it is constant

Since \(F_1\) and \(F_2\) share the identical \(\frac{\sqrt3}{2}\ln(x^2-xy+y^2)\) term, it cancels in the difference, leaving \[ c_1-c_2 = F_1-F_2=\arctan\frac{2y-x}{x\sqrt3}+\arctan\frac{2x-y}{y\sqrt3}=\arctan A+\arctan B, \] which is homogeneous of degree \(0\), i.e. a function of \(v=y/x\) only: \(A=(2v-1)/\sqrt3\), \(B=(2/v-1)/\sqrt3\). Sympy differentiates \(\arctan A+\arctan B\) with respect to \(v\) and simplifies the result to \(0\) identically — so \(c_1-c_2\) truly is constant on each of the two rays \(v>0\) and \(v<0\) (the value can only jump across \(v=0\), i.e. across \(x=0\) or \(y=0\), where \(A\) or \(B\) blows up).

4. Pin down the two constant values exactly

Sympy simplifies \[ A+B = \frac{2\sqrt3}{3}\cdot\frac{v^2-v+1}{v}, \qquad 1-AB = \frac{2}{3}\cdot\frac{v^2-v+1}{v}, \] so \(\dfrac{A+B}{1-AB}=\sqrt3=\tan\dfrac{\pi}{3}\) identically in \(v\) (the common factor \(v^2-v+1>0\) — it has negative discriminant — cancels and never vanishes). By the standard arctangent-addition identity \(\arctan A+\arctan B=\arctan\!\frac{A+B}{1-AB}+k\pi\), with \(k\) fixed by the sign of \(1-AB\), whose sign Sympy shows is exactly the sign of \(v\) (since \(v^2-v+1>0\) always):

• \(v=y/x>0\) (i.e. \(x,y\) same sign): \(1-AB>0\Rightarrow k=0\), so \(c_1-c_2=\arctan\sqrt3=\pi/3\). Numerically checked at \(v=0.1,0.5,1,2,5,10\): every value gives \(1.04719755\ldots=\pi/3\) to machine precision.
• \(v=y/x<0\) (i.e. \(x,y\) opposite signs): \(1-AB<0\) and \(A<0\), giving \(k=-1\), so \(c_1-c_2=\pi/3-\pi=-2\pi/3\). Numerically checked at \(v=-0.01,\ldots,-10\): every value gives \(-2.09439510\ldots=-2\pi/3\) to machine precision.

This resolves — with an actual proof rather than a guess — exactly the ambiguity the source's own step 9 flagged ("\(\pi/3\) or \(-2\pi/3\)") but never settled, and shows the source's separately boxed final answer in step 10, \(c_1=c_2+\pi\sqrt3/3\), is simply wrong: it has the wrong functional form (an extra factor of \(\sqrt3\) that does not belong — apparently carried over by mistake from the \(\sqrt3/2\) log coefficient — and is not even self-consistent with its own step 9).