Internal problem ID [8524]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 944.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [[_1st_order, _with_linear_symmetries], _rational, [_Abel, 2nd type, class C]]
Solve \begin {gather*} \boxed {y^{\prime }-\frac {-32 a y x -8 a^{2} x^{3}-16 x^{2} b a -32 a x +64 y^{3}+48 x^{2} a y^{2}+96 y^{2} b x +12 y a^{2} x^{4}+48 y a \,x^{3} b +48 b^{2} x^{2} y+a^{3} x^{6}+6 a^{2} x^{5} b +12 x^{4} b^{2} a +8 b^{3} x^{3}}{64 y+16 a \,x^{2}+32 x b +64}=0} \end {gather*}
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 47
dsolve(diff(y(x),x) = (-32*y(x)*a*x-8*a^2*x^3-16*a*x^2*b-32*a*x+64*y(x)^3+48*x^2*a*y(x)^2+96*y(x)^2*b*x+12*y(x)*a^2*x^4+48*y(x)*a*x^3*b+48*y(x)*b^2*x^2+a^3*x^6+6*a^2*x^5*b+12*a*x^4*b^2+8*b^3*x^3)/(64*y(x)+16*a*x^2+32*b*x+64),y(x), singsol=all)
\[ y \relax (x ) = -\frac {a \,x^{2}}{4}-\frac {b x}{2}+\RootOf \left (b x +2 \left (\int _{}^{\textit {\_Z}}-\frac {b \left (\textit {\_a} +1\right )}{2 \textit {\_a}^{3}+\textit {\_a} b +b}d \textit {\_a} \right )+2 c_{1}\right ) \]
✓ Solution by Mathematica
Time used: 1.02 (sec). Leaf size: 233
DSolve[y'[x] == (-32*a*x - 16*a*b*x^2 - 8*a^2*x^3 + 8*b^3*x^3 + 12*a*b^2*x^4 + 6*a^2*b*x^5 + a^3*x^6 - 32*a*x*y[x] + 48*b^2*x^2*y[x] + 48*a*b*x^3*y[x] + 12*a^2*x^4*y[x] + 96*b*x*y[x]^2 + 48*a*x^2*y[x]^2 + 64*y[x]^3)/(64 + 32*b*x + 16*a*x^2 + 64*y[x]),y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [x-4 \text {RootSum}\left [\text {$\#$1}^6 a^3+6 \text {$\#$1}^5 a^2 b+12 \text {$\#$1}^4 a^2 y(x)+12 \text {$\#$1}^4 a b^2+48 \text {$\#$1}^3 a b y(x)+8 \text {$\#$1}^3 b^3+8 \text {$\#$1}^2 a b+48 \text {$\#$1}^2 a y(x)^2+48 \text {$\#$1}^2 b^2 y(x)+16 \text {$\#$1} b^2+96 \text {$\#$1} b y(x)^2+32 b y(x)+32 b+64 y(x)^3\&,\frac {\text {$\#$1}^2 a \log (x-\text {$\#$1})+2 \text {$\#$1} b \log (x-\text {$\#$1})+4 y(x) \log (x-\text {$\#$1})+4 \log (x-\text {$\#$1})}{3 \text {$\#$1}^4 a^2+12 \text {$\#$1}^3 a b+24 \text {$\#$1}^2 a y(x)+12 \text {$\#$1}^2 b^2+48 \text {$\#$1} b y(x)+8 b+48 y(x)^2}\&\right ]=c_1,y(x)\right ] \]