Internal problem ID [8405]
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)]`]]
\[ \boxed {y^{\prime }-\sqrt {\frac {a y^{4}+b y^{2}+1}{a \,x^{4}+x^{2} b +1}}=0} \]
✓ Solution by Maple
Time used: 0.0 (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)
\[ \int _{}^{y \left (x \right )}\frac {1}{\sqrt {\textit {\_a}^{4} a +\textit {\_a}^{2} b +1}}d \textit {\_a} +\int _{}^{x}-\frac {\sqrt {\frac {a y \left (x \right )^{4}+b y \left (x \right )^{2}+1}{\textit {\_a}^{4} a +\textit {\_a}^{2} b +1}}}{\sqrt {a y \left (x \right )^{4}+b y \left (x \right )^{2}+1}}d \textit {\_a} +c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 46.307 (sec). Leaf size: 505
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 {$\#$1}^2 a+\sqrt {b^2-4 a}+b}{\sqrt {b^2-4 a}+b}} \sqrt {\frac {2 \text {$\#$1}^2 a}{b-\sqrt {b^2-4 a}}+1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a}{b+\sqrt {b^2-4 a}}} \text {$\#$1}\right ),\frac {b+\sqrt {b^2-4 a}}{b-\sqrt {b^2-4 a}}\right )}{\sqrt {2} \sqrt {\frac {a}{\sqrt {b^2-4 a}+b}} \sqrt {\text {$\#$1}^4 a+\text {$\#$1}^2 b+1}}\&\right ]\left [c_1-\frac {i \sqrt {\frac {\sqrt {b^2-4 a}+2 a x^2+b}{\sqrt {b^2-4 a}+b}} \sqrt {\frac {2 a x^2}{b-\sqrt {b^2-4 a}}+1} \operatorname {EllipticF}\left (i \text {arcsinh}\left (\sqrt {2} \sqrt {\frac {a}{b+\sqrt {b^2-4 a}}} x\right ),\frac {b+\sqrt {b^2-4 a}}{b-\sqrt {b^2-4 a}}\right )}{\sqrt {2} \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*}