Internal problem ID [8943]
Internal file name [OUTPUT/7878_Monday_June_06_2022_12_49_49_AM_3500330/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, Additional non-linear first order
Problem number: 609.
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.
\[ \boxed {y^{\prime }-\frac {-3 x^{2} y+F \left (x^{3} y\right )}{x^{3}}=0} \] Unable to determine ODE type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x^{3}+3 x^{2} y-F \left (x^{3} y\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-3 x^{2} y+F \left (x^{3} y\right )}{x^{3}} \end {array} \]
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 `, `-> Computing symmetries using: way = 5`[0, 1/x^3*F(y*x^3)]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 22
dsolve(diff(y(x),x) = (-3*x^2*y(x)+F(x^3*y(x)))/x^3,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\operatorname {RootOf}\left (x -\left (\int _{}^{\textit {\_Z}}\frac {1}{F \left (\textit {\_a} \right )}d \textit {\_a} \right )+c_{1} \right )}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.278 (sec). Leaf size: 117
DSolve[y'[x] == (F[x^3*y[x]] - 3*x^2*y[x])/x^3,y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^{y(x)}-\frac {x^3+F\left (x^3 K[2]\right ) \int _1^x\left (\frac {3 K[1]^5 K[2] F'\left (K[1]^3 K[2]\right )}{F\left (K[1]^3 K[2]\right )^2}-\frac {3 K[1]^2}{F\left (K[1]^3 K[2]\right )}\right )dK[1]}{F\left (x^3 K[2]\right )}dK[2]+\int _1^x\left (1-\frac {3 K[1]^2 y(x)}{F\left (K[1]^3 y(x)\right )}\right )dK[1]=c_1,y(x)\right ] \]