Solve the differential equation y = xy' + √(a^2 - b^2 y'^2)

Solve the differential equation $y = xy' + \sqrt{a^2 - b^2 y'^2}$.

Step 1: Recognize the Clairaut form

Writing \(p=y'\), the equation is \(y=xp+\sqrt{a^2-b^2p^2}=xp+f(p)\) with \(f(p)=\sqrt{a^2-b^2p^2}\). Since the right side depends on \(x\) only through the explicit term \(xp\), this is a genuine Clairaut equation.

Step 2: General solution

Replacing \(p\) by an arbitrary constant \(c\) gives the general solution \[ y = cx + \sqrt{a^2-b^2c^2}, \qquad |bc|\le|a|. \] Direct substitution (\texttt{sympy}, treating \(c\) as constant so \(y'=c\)) confirms this satisfies the original equation identically for every \(c\).

Step 3: Differentiate to find the envelope condition

Differentiating \(y=xp+\sqrt{a^2-b^2p^2}\) with respect to \(x\), with \(p=p(x)\), and using \(y'=p\): \[ p = p + p'(x)\Big(x-\frac{b^2p}{\sqrt{a^2-b^2p^2}}\Big) \ \Longrightarrow\ p'(x)\Big(x-\frac{b^2p}{\sqrt{a^2-b^2p^2}}\Big)=0. \] This was verified symbolically: expanding \(\dfrac{d}{dx}\big[xp+\sqrt{a^2-b^2p^2}\big]-p\) in \texttt{sympy} and comparing to \(p'(x)\big(x-\tfrac{b^2p}{\sqrt{a^2-b^2p^2}}\big)\) gives an identically zero difference, confirming the derivative step is correct. The singular solution comes from the second factor: \[ x = \frac{b^2p}{\sqrt{a^2-b^2p^2}}. \]

Step 4: Solve for \(p\)

Squaring, \(x^2(a^2-b^2p^2)=b^4p^2\), so \(a^2x^2=b^2p^2(b^2+x^2)\), giving \[ p=\pm\frac{ax}{b\sqrt{b^2+x^2}}. \] \texttt{sympy.solve} on \(x=\dfrac{b^2p}{\sqrt{a^2-b^2p^2}}\) returns exactly these two roots, confirming the branch \(p=\dfrac{ax}{b\sqrt{b^2+x^2}}\) (matching the sign of \(x\)) used next.

Step 5: Back-substitute for \(y\)

Substituting this \(p\) into \(y=xp+\sqrt{a^2-b^2p^2}\) and simplifying with \texttt{sympy} gives, after simplification, \[ y=\frac{a}{b}\sqrt{x^2+b^2}, \qquad\text{i.e.}\qquad b^2y^2=a^2(x^2+b^2). \]

Step 6: Direct verification in the original equation

To confirm this independently (not merely by trusting the elimination), \(y=\frac{a}{b}\sqrt{x^2+b^2}\) was differentiated directly: \(y'=\dfrac{ax}{b\sqrt{x^2+b^2}}\), and substituted into \(y-\big[xy'+\sqrt{a^2-b^2y'^2}\big]\); \texttt{sympy.simplify} reduces this residual to \(0\) identically. Implicit differentiation of \(b^2y^2=a^2(x^2+b^2)\) gives the same \(y'=\dfrac{a^2x}{b^2y}\), consistent with the explicit form. The singular solution is therefore confirmed to be \[ b^2y^2=a^2(x^2+b^2). \]

Step 7: Correct classification of the conic (source error)

Dividing \(b^2y^2-a^2x^2=a^2b^2\) by \(a^2b^2\) gives the standard form \[ \frac{y^2}{a^2}-\frac{x^2}{b^2}=1, \] verified term-by-term with \texttt{sympy}. This is the equation of a hyperbola for every choice of nonzero real \(a,b\) — the minus sign between the squared terms never produces an ellipse, regardless of whether \(ba\). The source's claim that the curve is "an ellipse or a hyperbola depending on whether \(ba\)" is therefore incorrect and is discarded. The source's alternate boxed form \(\dfrac{x^2}{a^2}+\dfrac{y^2}{a^2/b^2}=1\) was also checked against \(b^2y^2-a^2x^2-a^2b^2=0\) by clearing denominators in \texttt{sympy}; the two differ by \(a^2b^2+a^2x^2-a^2+x^2\ne0\) in general, confirming that alternate form contains a sign slip and must be dropped as well.