Internal problem ID [8920]
Internal file name [OUTPUT/7855_Monday_June_06_2022_12_47_34_AM_75020625/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 586.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(y)]`]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime }-\frac {F \left (\frac {y}{\sqrt {x^{2}+1}}\right ) x}{\sqrt {x^{2}+1}}=0} \] Unable to determine ODE type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } \sqrt {x^{2}+1}-F \left (\frac {y}{\sqrt {x^{2}+1}}\right ) x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {F \left (\frac {y}{\sqrt {x^{2}+1}}\right ) x}{\sqrt {x^{2}+1}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying homogeneous types: differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying an equivalence to an Abel ODE trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 3 `, `-> Computing symmetries using: way = 4`[(x^2+1)/x, y]
✓ Solution by Maple
Time used: 0.046 (sec). Leaf size: 64
dsolve(diff(y(x),x) = F(y(x)/(x^2+1)^(1/2))*x/(x^2+1)^(1/2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \operatorname {RootOf}\left (-F \left (\frac {\textit {\_Z}}{\sqrt {x^{2}+1}}\right ) \sqrt {x^{2}+1}+\textit {\_Z} \right ) \\ y \left (x \right ) &= \operatorname {RootOf}\left (-\ln \left (x^{2}+1\right )+2 \left (\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )-\textit {\_a}}d \textit {\_a} \right )+2 c_{1} \right ) \sqrt {x^{2}+1} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.933 (sec). Leaf size: 975
DSolve[y'[x] == (x*F[y[x]/Sqrt[1 + x^2]])/Sqrt[1 + x^2],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\left (-\frac {K[1] \sqrt {K[1]^2+1} F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )^3}{y(x) \left (K[1]^2 F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )^2-y(x)^2\right )}-\frac {K[1] F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )^2}{K[1]^2 F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )^2-y(x)^2}+\frac {K[1] F\left (\frac {y(x)}{\sqrt {K[1]^2+1}}\right )}{\sqrt {K[1]^2+1} y(x)}\right )dK[1]+\int _1^{y(x)}\left (-\frac {\sqrt {x^2+1} F\left (\frac {K[2]}{\sqrt {x^2+1}}\right )}{-x^2 F\left (\frac {K[2]}{\sqrt {x^2+1}}\right )^2-F\left (\frac {K[2]}{\sqrt {x^2+1}}\right )^2+K[2]^2}-\int _1^x\left (\frac {K[1] \sqrt {K[1]^2+1} \left (\frac {2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) K[1]^2}{\sqrt {K[1]^2+1}}-2 K[2]+\frac {2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )}{\sqrt {K[1]^2+1}}\right ) F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^3}{K[2] \left (K[1]^2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2-K[2]^2\right )^2}+\frac {K[1] \sqrt {K[1]^2+1} F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^3}{K[2]^2 \left (K[1]^2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2-K[2]^2\right )}-\frac {3 K[1] F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2}{K[2] \left (K[1]^2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2-K[2]^2\right )}+\frac {K[1] \left (\frac {2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) K[1]^2}{\sqrt {K[1]^2+1}}-2 K[2]+\frac {2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )}{\sqrt {K[1]^2+1}}\right ) F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2}{\left (K[1]^2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2-K[2]^2\right )^2}-\frac {2 K[1] F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right ) F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )}{\sqrt {K[1]^2+1} \left (K[1]^2 F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2+F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )^2-K[2]^2\right )}-\frac {K[1] F\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )}{\sqrt {K[1]^2+1} K[2]^2}+\frac {K[1] F'\left (\frac {K[2]}{\sqrt {K[1]^2+1}}\right )}{\left (K[1]^2+1\right ) K[2]}\right )dK[1]-\frac {K[2]}{-x^2 F\left (\frac {K[2]}{\sqrt {x^2+1}}\right )^2-F\left (\frac {K[2]}{\sqrt {x^2+1}}\right )^2+K[2]^2}\right )dK[2]=c_1,y(x)\right ] \]