4.42.15 $a^2{y}^{″}(x{)}^2={(y^{\prime }(x{)}^2+1)}^3$

ODE
$$a^2{y}^{″}(x{)}^2={(y^{\prime }(x{)}^2+1)}^3$$ ODE Classification

[[_2nd_order, _missing_x]]

Book solution method
TO DO

Mathematica ✓
cpu = 0.132494 (sec), leaf count = 141

$$\{\{y(x)\to c_2-i\sqrt{a^2(c_1^2-1)-2a{c}_{1}x+x^2}\},\{y(x)\to c_2+i\sqrt{a^2(c_1^2-1)-2a{c}_{1}x+x^2}\},\{y(x)\to c_2-i\sqrt{a^2(c_1^2-1)+2a{c}_{1}x+x^2}\},\{y(x)\to c_2+i\sqrt{a^2(c_1^2-1)+2a{c}_{1}x+x^2}\}\}$$

Maple ✓
cpu = 0.191 (sec), leaf count = 78

$$\{y\left(x\right)=-(x+\mathit{\_}\mathit{C}\mathit{1}+a)(-x-\mathit{\_}\mathit{C}\mathit{1}+a)\frac{1}{\sqrt{-{\mathit{\_}\mathit{C}\mathit{1}}^2-2\mathit{\_}\mathit{C}\mathit{1}x+a^2-x^2}}+\mathit{\_}\mathit{C}\mathit{2},y\left(x\right)=(x+\mathit{\_}\mathit{C}\mathit{1}+a)(-x-\mathit{\_}\mathit{C}\mathit{1}+a)\frac{1}{\sqrt{-{\mathit{\_}\mathit{C}\mathit{1}}^2-2\mathit{\_}\mathit{C}\mathit{1}x+a^2-x^2}}+\mathit{\_}\mathit{C}\mathit{2}\}$$ Mathematica raw input

DSolve[a^2*y''[x]^2 == (1 + y'[x]^2)^3,y[x],x]

Mathematica raw output

{{y[x] -> (-I)*Sqrt[x^2 - 2*a*x*C[1] + a^2*(-1 + C[1]^2)] + C[2]}, {y[x] -> I*Sq
rt[x^2 - 2*a*x*C[1] + a^2*(-1 + C[1]^2)] + C[2]}, {y[x] -> (-I)*Sqrt[x^2 + 2*a*x
*C[1] + a^2*(-1 + C[1]^2)] + C[2]}, {y[x] -> I*Sqrt[x^2 + 2*a*x*C[1] + a^2*(-1 +
 C[1]^2)] + C[2]}}

Maple raw input

dsolve(a^2*diff(diff(y(x),x),x)^2 = (1+diff(y(x),x)^2)^3, y(x),'implicit')

Maple raw output

y(x) = -(x+_C1+a)*(-x-_C1+a)/(-_C1^2-2*_C1*x+a^2-x^2)^(1/2)+_C2, y(x) = (x+_C1+a
)*(-x-_C1+a)/(-_C1^2-2*_C1*x+a^2-x^2)^(1/2)+_C2