Solve the differential equation y(y - xy') = √(x^4 + y^4)

Solve the differential equation $y(y - xy') = \sqrt{x^4 + y^4}$.

Expand and recognize homogeneity

Expanding gives \(y^2 - xyy' = \sqrt{x^4+y^4}\). Since \(\sqrt{x^4+y^4}=x^2\sqrt{1+(y/x)^4}\), the equation is homogeneous of degree 2 in \(x,y\).

Substitute \(y=vx\)

With \(y=vx\) and \(y'=v+xv'\), the left side becomes \(vx\big(vx-x(v+xv')\big)=-vx^3v'\), and the right side becomes \(x^2\sqrt{1+v^4}\). Dividing by \(x^2\) (for \(x\\neq0\)) gives\n\\\[ -vx\\,\\frac{dv}{dx} = \\sqrt{1+v^4}. \\\]

Separate variables

\[ \\frac{v\\,dv}{\\sqrt{1+v^4}} = -\\frac{dx}{x}. \]

Integrate both sides

With \(u=v^2\), \(du=2v\\,dv\), so the left integral becomes \(\\tfrac12\\int \\frac{du}{\\sqrt{1+u^2}} = \\tfrac12\\ln\\!\\big(u+\\sqrt{1+u^2}\\big)\). Integrating the right side gives \(-\\ln|x|+C_1\). Multiplying by 2 and exponentiating,\n\\\[ v^2+\\sqrt{1+v^4} = \\frac{K}{x^2}, \\\]\nwith \(K>0\) an arbitrary constant.

Back-substitute \(v=y/x\)

Multiplying through by \(x^2\) and simplifying \(x^2\\sqrt{1+y^4/x^4}=\\sqrt{x^4+y^4}\) gives the implicit solution \(y^2+\\sqrt{x^4+y^4}=C\).

Independent verification

Differentiating \(y^2+\\sqrt{x^4+y^4}=C\) implicitly gives \(y' = -\\dfrac{x^3}{y\\big(y^2+\\sqrt{x^4+y^4}\\big)}\). Substituting this back into \(y(y-xy')\) and simplifying with a computer algebra system shows it collapses identically to \(\\sqrt{x^4+y^4}\) for all \(x,y\) — confirming the solution exactly, with no leftover residual.