Find the condition under which P dx+Q dy=0 admits an…

Find the condition under which \(P\,dx+Q\,dy=0\) admits an integrating factor of the form \(\mu=\mu(y^2-x)\). Show that this factor can be obtained by quadrature. Apply this to the case \[ P = (2y^2-2x-1)e^x+(2y^2-3x)e^y,\qquad Q = 2ye^x+2x(y^2+y-x)e^y. \]

1. General condition (Part 1) — verified

With \(u=y^2-x\) (\(u_x=-1,u_y=2y\)), Sympy differentiation of \(\mu(u)P\) and \(\mu(u)Q\) and setting them equal gives \[ \mu'(u)(2yP+Q) = \mu(u)\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right) \;\Longrightarrow\; \boxed{\frac{\partial Q/\partial x-\partial P/\partial y}{2yP+Q} = f(y^2-x)} \] for some function \(f\), and then \(\mu(u)=\exp\big(\int f(u)\,du\big)\) by quadrature — both confirmed correct and matching the source.

2. Recompute \(\partial P/\partial y\) and \(\partial Q/\partial x\) independently

Sympy differentiation of \(P=(2y^2-2x-1)e^x+(2y^2-3x)e^y\) gives, by the product rule on the second term, \[ \frac{\partial P}{\partial y} = 4y\,e^x + (4y+2y^2-3x)\,e^y, \] whereas the source's step 5 gives only \(4ye^x+4ye^y\) — the \((2y^2-3x)e^y\) contribution from \(\partial_y[(2y^2-3x)e^y]\) was dropped. (\(\partial Q/\partial x\) was computed correctly by the source: Sympy confirms \(\partial Q/\partial x = 2ye^x+(2y^2+2y-4x)e^y\).)

3. Recompute the ratio with the corrected derivative

With the correct \(\partial P/\partial y\), Sympy gives \[ \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} = -2ye^x - (x+2y)e^y, \] (genuinely different from the source's \(-2ye^x+(2y^2-2y-4x)e^y\)), and, computing \(2yP+Q\) and simplifying the full ratio, \[ \frac{\partial Q/\partial x-\partial P/\partial y}{2yP+Q} = \frac{1}{2(x-y^2)}. \] Substituting \(x=y^2-u\) to express this purely in terms of \(u=y^2-x\) (and \(y\), which must cancel for the condition to hold — Sympy confirms it does) gives exactly \[ f(u) = -\frac{1}{2u}, \] not the constant \(1\) the source asserted with no derivation.

4. Find \(\mu\) by quadrature and verify

\(\mu(u)=\exp\!\big(\int -\tfrac{1}{2u}\,du\big) = \exp(-\tfrac12\ln|u|) = |u|^{-1/2}\), so \[ \boxed{\mu = (y^2-x)^{-1/2}}. \] Sympy confirms this directly: multiplying \(P,Q\) by \(\mu=(y^2-x)^{-1/2}\) and computing \(\partial(\mu P)/\partial y-\partial(\mu Q)/\partial x\) gives \(0\) identically. By contrast, repeating the same exactness check with the source's claimed \(\mu=e^{y^2-x}\) gives a nonzero residual — a direct, independent proof that the source's Part 2 answer is wrong.