Solve the differential equation y^2 + x^2(y')^5 =…
Solve the differential equation $y^2 + x^2(y')^5 = xy\big((y')^2+(y')^3\big)$.
Explanation of Method
Writing \(p=y'\), the equation \(y^2+x^2p^5-xyp^2-xyp^3=0\) turns out to factor algebraically over \(x,y,p\) into two much lower-degree pieces, \((xp^2-y)(xp^3-y)=0\), so the problem reduces to solving two simple separable first-order equations rather than a genuine quintic.
Set \(p=y'\) and rearrange
\[y^2+x^2p^5-xyp^2-xyp^3=0.\] Taken at face value this is a degree-5 polynomial in \(p\), which is why the source concluded the problem “may involve special functions.” That conclusion is checked, not assumed.
Factor the expression directly
Remarkably, the left side factors exactly: \[y^2+x^2p^5-xyp^2-xyp^3=(xp^2-y)(xp^3-y).\] (Confirmed by symbolic expansion: multiplying out reproduces the left side term for term.) So the fifth-degree equation is equivalent to the pair \[xp^2=y \quad\text{or}\quad xp^3=y,\] i.e. two genuinely simple first-order ODEs, not one irreducible quintic.
Solve Case A: \(x(y')^2=y\)
\(y'=\pm\sqrt{y/x}\) is separable: \(dy/\sqrt y=\pm dx/\sqrt x\), so \(2\sqrt y=\pm2\sqrt x+\text{const}\), giving \[y=(\sqrt x+C)^2\] (the constant's sign absorbs the \(\pm\)). Substituting back into \(x(y')^2=y\) confirms this holds identically for every \(C\).
Solve Case B: \(x(y')^3=y\)
Using the real cube root, \(y'=(y/x)^{1/3}\) is separable: \(y^{-1/3}dy=x^{-1/3}dx\), so \(\tfrac32y^{2/3}=\tfrac32x^{2/3}+\text{const}\), giving the implicit family \(y^{2/3}=x^{2/3}+C\), equivalently \[y=\pm\left(x^{2/3}+C\right)^{3/2}.\] Substituting back into \(x(y')^3=y\) confirms this holds identically for every \(C\) and either sign.
Recover the line solutions as special members
Setting \(C=0\) in Case A gives \(y=x\); setting \(C=0\) with the minus sign in Case B gives \(y=-x\); and \(y\equiv0\) is a common degenerate solution of both factors (lost during the division by \(y\) or \(w\) in separation, so it must be listed separately). All three were verified directly against the original equation and match the source's step 8 computation, but they now appear as identified special cases of two explicit general families rather than as the entire (incomplete) answer.
Result
Final answer:
Corrected/expanded. The source stopped after finding only the three line solutions and declared the general problem intractable. In fact the quintic factors cleanly as \((xy'^2-y)(xy'^3-y)=0\), giving two complete explicit solution families:
\[\boxed{\,x(y')^2=y:\quad y=(\sqrt{x}+C)^2\,}\]
\[\boxed{\,x(y')^3=y:\quad y=\pm\left(x^{2/3}+C\right)^{3/2}\,}\]
together with the degenerate common solution \(y=0\); the previously found lines \(y=x\) (Case A, \(C=0\)) and \(y=-x\) (Case B, \(C=0\), minus sign) are special members of these families. Every branch was confirmed by direct symbolic substitution into the original equation.