Internal problem ID [7649]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 68.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [[_1st_order, _with_symmetry_[F(x),G(x)*y+H(x)]]]
Solve \begin {gather*} \boxed {y^{\prime }-\sqrt {\frac {a y^{4}+b y^{2}+1}{a \,x^{4}+x^{2} b +1}}=0} \end {gather*}
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 77
dsolve(diff(y(x),x) - sqrt((a*y(x)^4+b*y(x)^2+1)/(a*x^4+b*x^2+1))=0,y(x), singsol=all)
\[ \int _{}^{y \relax (x )}\frac {1}{\sqrt {a \,\textit {\_a}^{4}+\textit {\_a}^{2} b +1}}d \textit {\_a} +\int _{}^{x}-\frac {\sqrt {\frac {y \relax (x )^{4} a +b y \relax (x )^{2}+1}{a \,\textit {\_a}^{4}+\textit {\_a}^{2} b +1}}}{\sqrt {y \relax (x )^{4} a +b y \relax (x )^{2}+1}}d \textit {\_a} +c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 50.537 (sec). Leaf size: 475
DSolve[y'[x] - Sqrt[(a*y[x]^4+b*y[x]^2+1)/(a*x^4+b*x^2+1)]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {i \sqrt {\frac {2 \text {$\#$1}^2 a+\sqrt {b^2-4 a}+b}{\sqrt {b^2-4 a}+b}} \sqrt {\frac {2 \text {$\#$1}^2 a}{b-\sqrt {b^2-4 a}}+1} \text {EllipticF}\left (i \sinh ^{-1}\left (\sqrt {2} \text {$\#$1} \sqrt {\frac {a}{\sqrt {b^2-4 a}+b}}\right ),\frac {\sqrt {b^2-4 a}+b}{b-\sqrt {b^2-4 a}}\right )}{\sqrt {2} \sqrt {\frac {a}{\sqrt {b^2-4 a}+b}} \sqrt {\text {$\#$1}^4 a+\text {$\#$1}^2 b+1}}\&\right ]\left [c_1-\frac {i \sqrt {2 x^2 \left (b-\sqrt {b^2-4 a}\right )+4} \sqrt {x^2 \left (\sqrt {b^2-4 a}+b\right )+2} \text {EllipticF}\left (i \sinh ^{-1}\left (\sqrt {2} x \sqrt {\frac {a}{\sqrt {b^2-4 a}+b}}\right ),\frac {b \left (\sqrt {b^2-4 a}+b\right )}{2 a}-1\right )}{4 \sqrt {\frac {a}{\sqrt {b^2-4 a}+b}} \sqrt {a x^4+b x^2+1}}\right ] \\ y(x)\to -\frac {\sqrt {-\frac {\sqrt {b^2-4 a}+b}{a}}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {-\frac {\sqrt {b^2-4 a}+b}{a}}}{\sqrt {2}} \\ y(x)\to -\frac {\sqrt {\frac {\sqrt {b^2-4 a}-b}{a}}}{\sqrt {2}} \\ y(x)\to \frac {\sqrt {\frac {\sqrt {b^2-4 a}-b}{a}}}{\sqrt {2}} \\ \end{align*}