4.24.22 $f(y(x)y^{\prime }(x)+x)=y(x{)}^2(y^{\prime }(x{)}^2+1)$

ODE
$$f(y(x)y^{\prime }(x)+x)=y(x{)}^2(y^{\prime }(x{)}^2+1)$$ ODE Classification

[`x=_G(y,y')`]

Book solution method
Change of variable

Mathematica ✓
cpu = 0.0989407 (sec), leaf count = 42

$$\text{Solve}[\{\text{K}\mathrm{\$}\text{278007}y(\text{K}\mathrm{\$}\text{278007})+x=f^{(-1)}(({\text{K}\mathrm{\$}\text{278007}}^2+1)y(\text{K}\mathrm{\$}\text{278007}{)}^2),y(x)=\frac{c_1}{\sqrt{{\text{K}\mathrm{\$}\text{278007}}^2+1}}\},\{y(x),\text{K}\mathrm{\$}\text{278007}\}]$$

Maple ✓
cpu = 0.151 (sec), leaf count = 47

$$\{[x\left(\mathit{\_}\mathit{T}\right)=1(\mathit{R}\mathit{o}\mathit{o}\mathit{t}\mathit{O}\mathit{f}({\mathit{\_}\mathit{C}\mathit{1}}^2-f\left(\mathit{\_}\mathit{Z}\right))\sqrt{{\mathit{\_}\mathit{T}}^2+1}-\mathit{\_}\mathit{C}\mathit{1}\mathit{\_}\mathit{T})\frac{1}{\sqrt{{\mathit{\_}\mathit{T}}^2+1}},y\left(\mathit{\_}\mathit{T}\right)=\mathit{\_}\mathit{C}\mathit{1}\frac{1}{\sqrt{{\mathit{\_}\mathit{T}}^2+1}}]\}$$ Mathematica raw input

DSolve[f[x + y[x]*y'[x]] == y[x]^2*(1 + y'[x]^2),y[x],x]

Mathematica raw output

Solve[{x + K$278007*y[K$278007] == InverseFunction[f, 1, 1][(1 + K$278007^2)*y[K
$278007]^2], y[x] == C[1]/Sqrt[1 + K$278007^2]}, {y[x], K$278007}]

Maple raw input

dsolve(f(y(x)*diff(y(x),x)+x) = (1+diff(y(x),x)^2)*y(x)^2, y(x),'implicit')

Maple raw output

[x(_T) = (RootOf(_C1^2-f(_Z))*(_T^2+1)^(1/2)-_C1*_T)/(_T^2+1)^(1/2), y(_T) = _C1
/(_T^2+1)^(1/2)]