Internal problem ID [3453]
Internal file name [OUTPUT/2946_Sunday_June_05_2022_08_47_49_AM_80337687/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 7
Problem number: 197.
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.
\[ \boxed {x y^{\prime }-y+x \left (x -y\right ) \sqrt {x^{2}+y^{2}}=0} \] Unable to determine ODE type.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-y+x \left (x -y\right ) \sqrt {x^{2}+y^{2}}=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y-x \left (x -y\right ) \sqrt {x^{2}+y^{2}}}{x} \end {array} \]
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: 0.078 (sec). Leaf size: 50
dsolve(x*diff(y(x),x) = y(x)-x*(x-y(x))*sqrt(x^2+y(x)^2),y(x), singsol=all)
\[ \ln \left (2\right )+\ln \left (\frac {x \left (\sqrt {2 x^{2}+2 y \left (x \right )^{2}}+y \left (x \right )+x \right )}{-x +y \left (x \right )}\right )+\frac {\sqrt {2}\, x^{2}}{2}-\ln \left (x \right )-c_{1} = 0 \]
✓ Solution by Mathematica
Time used: 1.529 (sec). Leaf size: 84
DSolve[x y'[x]==y[x]-x(x-y[x])Sqrt[x^2+y[x]^2],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*}