Solve 2(x^2y + √(1+x^4y^2))dx + x^3 dy = 0
Solve $2\left(x^2y + \sqrt{1+x^4y^2}\right)dx + x^3\,dy = 0$.
Explanation of Method
The equation is not exact, but the combination \(x^4y^2\) under the radical suggests the substitution \(u=x^2y\), which collapses the equation to a separable one in \(u\) and \(x\).
Check exactness
With \(M = 2x^2y+2\\sqrt{1+x^4y^2}\), \(N=x^3\): \(M_y = 2x^2 + \\dfrac{2x^4y}{\\sqrt{1+x^4y^2}} \\neq N_x = 3x^2\), so the equation is not exact.
Substitute \(u = x^2y\)
Then \(du = 2xy\\,dx + x^2\\,dy\), so \(x^2\\,dy = du - 2xy\\,dx\). Substituting into \(2x^2y\\,dx+2\\sqrt{1+u^2}\\,dx+x^3dy=0\), the \(2x^2y\\,dx\) terms cancel, leaving \(2\\sqrt{1+u^2}\\,dx + x\\,du = 0\).
Separate and integrate
\(\\dfrac{du}{\\sqrt{1+u^2}} = -\\dfrac{2\\,dx}{x}\), so \(\\sinh^{-1}(u) = -2\\ln|x| + C\), i.e. \(u+\\sqrt{1+u^2} = K/x^2\).
Back-substitute \(u=x^2y\)
\[ x^2y + \\sqrt{1+x^4y^2} = \\frac{C}{x^2}. \]
Solve for \(y\) explicitly
Isolating the radical, squaring, and simplifying: \(1 = C^2/x^4 - 2Cy\), so \[ y = \\frac{C^2-x^4}{2Cx^4}. \] Substituting this back into \(x^2y+\\sqrt{1+x^4y^2}\) and simplifying with sympy (taking \(C,x>0\)) reduces identically to \(C/x^2\), confirming the explicit form is consistent with the implicit one.
Verify against the original equation
Implicit-differentiating \(F(x,y)=x^2y+\\sqrt{1+x^4y^2}-C/x^2=0\) with sympy and solving for \(dy/dx\) gives an expression that, after eliminating \(C\) using \(C=x^4y+x^2\\sqrt{1+x^4y^2}\) (from \(F=0\)), simplifies exactly to \(-2(x^2y+\\sqrt{1+x^4y^2})/x^3\) — precisely the slope required by the original equation \(2(x^2y+\\sqrt{1+x^4y^2})dx+x^3dy=0\). The residual simplifies to \(0\) identically. \(\\checkmark\)
Result
Final answer:
Boxed result: \[ \\boxed{x^2y + \\sqrt{1+x^4y^2} = \\frac{C}{x^2}} \] or equivalently \(x^4y + x^2\\sqrt{1+x^4y^2}=C\), with explicit form \(y = \\dfrac{C^2-x^4}{2Cx^4}\).