Internal problem ID [8452]
Internal file name [OUTPUT/7385_Sunday_June_05_2022_10_54_10_PM_34418168/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 115.
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(x)*y+H(x)]`]]
Unable to solve or complete the solution.
\[ \boxed {x y^{\prime }-x \left (y-x \right ) \sqrt {y^{2}+x^{2}}-y=0} \] Unable to determine ODE type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-x \left (y-x \right ) \sqrt {y^{2}+x^{2}}-y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x \left (y-x \right ) \sqrt {y^{2}+x^{2}}+y}{x} \end {array} \]
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 = 5`[0, (x-y)*(x^2+y^2)^(1/2)/x]
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 50
dsolve(x*diff(y(x),x) - x*(y(x)-x)*sqrt(y(x)^2 + x^2) - y(x)=0,y(x), singsol=all)
\[ \ln \left (2\right )+\ln \left (\frac {x \left (\sqrt {2 y \left (x \right )^{2}+2 x^{2}}+y \left (x \right )+x \right )}{y \left (x \right )-x}\right )+\frac {\sqrt {2}\, x^{2}}{2}-\ln \left (x \right )-c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 1.405 (sec). Leaf size: 84
DSolve[x*y'[x] - x*(y[x]-x)*Sqrt[y[x]^2 + x^2] - y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x \tanh \left (\frac {x^2+2 c_1}{2 \sqrt {2}}\right ) \left (2+\sqrt {2} \tanh \left (\frac {x^2+2 c_1}{2 \sqrt {2}}\right )\right )}{\sqrt {2}+2 \tanh \left (\frac {x^2+2 c_1}{2 \sqrt {2}}\right )} \\ y(x)\to x \\ \end{align*}