60.1.68 problem 68
Internal
problem
ID
[10082]
Book
:
Differential
Gleichungen,
E.
Kamke,
3rd
ed.
Chelsea
Pub.
NY,
1948
Section
:
Chapter
1,
linear
first
order
Problem
number
:
68
Date
solved
:
Monday, January 27, 2025 at 06:24:59 PM
CAS
classification
:
[[_1st_order, `_with_symmetry_[F(x),G(x)*y+H(x)]`]]
\begin{align*} y^{\prime }-\sqrt {\frac {a y^{4}+b y^{2}+1}{a \,x^{4}+b \,x^{2}+1}}&=0 \end{align*}
✓ Solution by Maple
Time used: 0.011 (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)
\[
\int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} a +b \,\textit {\_a}^{2}+1}}d \textit {\_a} -\frac {\int _{}^{x}\sqrt {\frac {a y^{4}+b y^{2}+1}{\textit {\_a}^{4} a +b \,\textit {\_a}^{2}+1}}d \textit {\_a}}{\sqrt {a y^{4}+b y^{2}+1}}+c_{1} = 0
\]
✓ Solution by Mathematica
Time used: 45.858 (sec). Leaf size: 505
DSolve[D[y[x],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*}