Given the differential equation (1)/(x)(λ x^2 - y^2) dx + (μ…

Given the differential equation \[ \frac{1}{x}(\lambda x^2 - y^2)\,dx + (\mu x + 2\lambda y)\,dy = 0, \] \(1^{\circ}\) determine the parameters \(\lambda\) and \(\mu\) such that the equation has an integrating factor \(\dfrac{1}{x^2+y^2}\). \(2^{\circ}\) For the values of \(\lambda,\mu\) thus obtained, find the particular solution \(\varphi(x,y)=0\) satisfying \(\varphi(a,0)=0\) \((a>0)\).

1. Rederive the exactness condition from scratch

With \(M=(\lambda x^2-y^2)/x\), \(N=\mu x+2\lambda y\), \(\nu=1/(x^2+y^2)\), Sympy computes \(\partial(\nu M)/\partial y-\partial(\nu N)/\partial x\), clears the common denominator \((x^2+y^2)^2\), and simplifies the numerator to \[ 2\lambda xy + \mu x^2 - \mu y^2 - 2xy \;=\; 2xy(\lambda-1) - \mu(x^2-y^2) \;=\; 0, \] i.e. \(2xy(\lambda-1)=\mu(y^2-x^2)\) — matching the source's condition exactly, so this part of the source's work is correct.

2. Solve the condition rigorously as a polynomial identity

The condition must hold for every \((x,y)\), so it is a polynomial identity in the independent variables \(x,y\): \(2(\lambda-1)\,xy + \mu\,x^2 - \mu\,y^2 = 0\) for all \(x,y\). Collecting by Sympy's \(\texttt{Poly}(\cdot,x,y)\) gives three independent coefficients — of \(xy\), \(x^2\), \(y^2\) — that must each vanish: \[ 2(\lambda-1)=0,\qquad \mu = 0,\qquad -\mu=0. \] Solving this system with Sympy (\(\texttt{solve}\)) gives the unique solution \[ \boxed{\lambda = 1,\quad \mu = 0}. \] This is exactly the value the source's own (correct) partial coefficient-matching found for \(\mu\) before it was discarded in favor of the unjustified \(\mu=2\); \(\lambda=1,\mu=2\) does not solve \(2xy(\lambda-1)=\mu(y^2-x^2)\) (it gives \(0=2(y^2-x^2)\), false for \(x\neq\pm y\)), so the source's boxed answer is confirmed wrong.

3. Find the true general solution with \(\lambda=1,\mu=0\)

The equation becomes \(\dfrac{x^2-y^2}{x}\,dx+2y\,dy=0\), i.e. \(M=(x^2-y^2)/x\), \(N=2y\). Multiplying by \(\nu=1/(x^2+y^2)\) gives \(\tilde M=\dfrac{x^2-y^2}{x(x^2+y^2)}\), \(\tilde N=\dfrac{2y}{x^2+y^2}\); Sympy confirms \(\partial\tilde M/\partial y-\partial\tilde N/\partial x=0\) exactly. Integrating \(\tilde N\) in \(y\) gives \(\ln(x^2+y^2)+g(x)\); matching \(\partial F/\partial x=\tilde M\) forces \(g'(x)=-1/x\), so \[ F(x,y) = \ln(x^2+y^2) - \ln|x| = \ln\frac{x^2+y^2}{x}. \] Sympy confirms this satisfies both partials exactly. The general integral \(F=C\) is equivalent to \[ x^2+y^2 = Kx, \] the classical one-parameter family of circles through the origin centered on the \(x\)-axis. (Implicit differentiation of \(x^2+y^2=Kx\) reproduces \(y'=(y^2-x^2)/(2xy)\), matching \(-M/N\) from the original equation exactly — verified with Sympy.) This is a genuinely different, and much simpler, family than the source's \(\ln|x|+\arctan(y/x)=C\), which was built on the incorrect \(\lambda=1,\mu=2\).

4. Apply \(\varphi(a,0)=0\)

At \((a,0)\): \(a^2+0=Ka\Rightarrow K=a\). The particular solution is \[ x^2+y^2 = ax \quad\Longleftrightarrow\quad \left(x-\frac a2\right)^2+y^2=\left(\frac a2\right)^2, \] the circle of radius \(a/2\) centered at \((a/2,0)\), passing through the origin and through \((a,0)\).