ODE
\[ 3 y'(x)^5-y(x) y'(x)+1=0 \] ODE Classification
[_quadrature]
Book solution method
Missing Variables ODE, Independent variable missing, Solve for \(y\)
Mathematica ✓
cpu = 0.454499 (sec), leaf count = 171
\[\left \{\text {Solve}\left [c_1+x=\int _1^{y(x)} \frac {1}{\text {Root}\left [-\text {$\#$1} K[1]+3 \text {$\#$1}^5+1\& ,1\right ]} \, dK[1],y(x)\right ],\text {Solve}\left [c_1+x=\int _1^{y(x)} \frac {1}{\text {Root}\left [-\text {$\#$1} K[2]+3 \text {$\#$1}^5+1\& ,2\right ]} \, dK[2],y(x)\right ],\text {Solve}\left [c_1+x=\int _1^{y(x)} \frac {1}{\text {Root}\left [-\text {$\#$1} K[3]+3 \text {$\#$1}^5+1\& ,3\right ]} \, dK[3],y(x)\right ],\text {Solve}\left [c_1+x=\int _1^{y(x)} \frac {1}{\text {Root}\left [-\text {$\#$1} K[4]+3 \text {$\#$1}^5+1\& ,4\right ]} \, dK[4],y(x)\right ],\text {Solve}\left [c_1+x=\int _1^{y(x)} \frac {1}{\text {Root}\left [-\text {$\#$1} K[5]+3 \text {$\#$1}^5+1\& ,5\right ]} \, dK[5],y(x)\right ]\right \}\]
Maple ✓
cpu = 0.174 (sec), leaf count = 50
\[ \left \{ x-{\frac { \left ( y \left ( x \right ) \right ) ^{2}}{2}}-{\frac {5\, \left ( {\it RootOf} \left ( 3\,{{\it \_Z}}^{5}-y \left ( x \right ) {\it \_Z}+1 \right ) \right ) ^{3}}{2}}+{\frac {3\,y \left ( x \right ) \left ( {\it RootOf} \left ( 3\,{{\it \_Z}}^{5}-y \left ( x \right ) {\it \_Z}+1 \right ) \right ) ^{4}}{2}}-{\it \_C1}=0 \right \} \] Mathematica raw input
DSolve[1 - y[x]*y'[x] + 3*y'[x]^5 == 0,y[x],x]
Mathematica raw output
{Solve[x + C[1] == Integrate[Root[1 - K[1]*#1 + 3*#1^5 & , 1]^(-1), {K[1], 1, y[x]}], y[x]], Solve[x + C[1] == Integrate[Root[1 - K[2]*#1 + 3*#1^5 & , 2]^(-1), {K[2], 1, y[x]}], y[x]], Solve[x + C[1] == Integrate[Root[1 - K[3]*#1 + 3*#1^5 & , 3]^(-1), {K[3], 1, y[x]}], y[x]], Solve[x + C[1] == Integrate[Root[1 - K[4]*#1 + 3*#1^5 & , 4]^(-1), {K[4], 1, y[x]}], y[x]], Solve[x + C[1] == Integrate[Root[1 - K[5]*#1 + 3*#1^5 & , 5]^(-1), {K[5], 1, y[x]}], y[x]]}
Maple raw input
dsolve(3*diff(y(x),x)^5-y(x)*diff(y(x),x)+1 = 0, y(x),'implicit')
Maple raw output
x-1/2*y(x)^2-5/2*RootOf(3*_Z^5-y(x)*_Z+1)^3+3/2*y(x)*RootOf(3*_Z^5-y(x)*_Z+1)^4-_C1 = 0