ODE
\[ 2 y(x) y''(x)=y'(x)^2 \left (y'(x)^2+1\right ) \] ODE Classification
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
Book solution method
TO DO
Mathematica ✓
cpu = 0.908026 (sec), leaf count = 173
\[\left \{\left \{y(x)\to \text {InverseFunction}\left [-i e^{-c_1} \left (\sqrt {\text {$\#$1}} \sqrt {\text {$\#$1} e^{2 c_1}-1}-e^{-c_1} \log \left (\sqrt {\text {$\#$1}} e^{2 c_1}+e^{c_1} \sqrt {\text {$\#$1} e^{2 c_1}-1}\right )\right )\& \right ]\left [c_2+x\right ]\right \},\left \{y(x)\to \text {InverseFunction}\left [i e^{-c_1} \left (\sqrt {\text {$\#$1}} \sqrt {\text {$\#$1} e^{2 c_1}-1}-e^{-c_1} \log \left (\sqrt {\text {$\#$1}} e^{2 c_1}+e^{c_1} \sqrt {\text {$\#$1} e^{2 c_1}-1}\right )\right )\& \right ]\left [c_2+x\right ]\right \}\right \}\]
Maple ✓
cpu = 0.078 (sec), leaf count = 87
\[ \left \{ -{\frac {{\it \_C1}}{2}\arctan \left ( {1 \left ( y \left ( x \right ) -{\frac {{\it \_C1}}{2}} \right ) {\frac {1}{\sqrt {y \left ( x \right ) \left ( {\it \_C1}-y \left ( x \right ) \right ) }}}} \right ) }-\sqrt {y \left ( x \right ) \left ( {\it \_C1}-y \left ( x \right ) \right ) }-x-{\it \_C2}=0,{\frac {{\it \_C1}}{2}\arctan \left ( {1 \left ( y \left ( x \right ) -{\frac {{\it \_C1}}{2}} \right ) {\frac {1}{\sqrt {y \left ( x \right ) \left ( {\it \_C1}-y \left ( x \right ) \right ) }}}} \right ) }+\sqrt {y \left ( x \right ) \left ( {\it \_C1}-y \left ( x \right ) \right ) }-x-{\it \_C2}=0 \right \} \] Mathematica raw input
DSolve[2*y[x]*y''[x] == y'[x]^2*(1 + y'[x]^2),y[x],x]
Mathematica raw output
{{y[x] -> InverseFunction[((-I)*(-(Log[E^(2*C[1])*Sqrt[#1] + E^C[1]*Sqrt[-1 + E^(2*C[1])*#1]]/E^C[1]) + Sqrt[#1]*Sqrt[-1 + E^(2*C[1])*#1]))/E^C[1] & ][x + C[2]]}, {y[x] -> InverseFunction[(I*(-(Log[E^(2*C[1])*Sqrt[#1] + E^C[1]*Sqrt[-1 + E^(2*C[1])*#1]]/E^C[1]) + Sqrt[#1]*Sqrt[-1 + E^(2*C[1])*#1]))/E^C[1] & ][x + C[2]]}}
Maple raw input
dsolve(2*y(x)*diff(diff(y(x),x),x) = diff(y(x),x)^2*(1+diff(y(x),x)^2), y(x),'implicit')
Maple raw output
-1/2*_C1*arctan((y(x)-1/2*_C1)/(y(x)*(_C1-y(x)))^(1/2))-(y(x)*(_C1-y(x)))^(1/2)-x-_C2 = 0, 1/2*_C1*arctan((y(x)-1/2*_C1)/(y(x)*(_C1-y(x)))^(1/2))+(y(x)*(_C1-y(x)))^(1/2)-x-_C2 = 0