Internal problem ID [10090]
Internal file name [OUTPUT/9037_Monday_June_06_2022_06_15_52_AM_89510699/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1769 (book 6.178).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is"
Maple gives the following as the ode type
[[_2nd_order, _exact, _nonlinear], [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_y_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {x \left (x +y\right ) y^{\prime \prime }+x {y^{\prime }}^{2}+\left (x -y\right ) y^{\prime }-y=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x y+x^{2}\right ) y^{\prime \prime }+\left (\left (y^{\prime }+1\right ) x -y\right ) y^{\prime }-y\right )d x &= 0 \\ -x y-y^{2}+\left (x y+x^{2}\right ) y^{\prime } = c_{1} \end {align*}
Which is now solved for \(y\). Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} -x^{2} u \left (x \right )-u \left (x \right )^{2} x^{2}+\left (x^{2} u \left (x \right )+x^{2}\right ) \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) = c_{1} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {c_{1}}{x^{3} \left (u +1\right )} \end {align*}
Where \(f(x)=\frac {c_{1}}{x^{3}}\) and \(g(u)=\frac {1}{u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{u +1}} \,du &= \frac {c_{1}}{x^{3}} \,d x \\ \int { \frac {1}{\frac {1}{u +1}} \,du} &= \int {\frac {c_{1}}{x^{3}} \,d x} \\ \frac {u \left (u +2\right )}{2}&=-\frac {c_{1}}{2 x^{2}}+c_{3} \\ \end{align*} The solution is \[ \frac {u \left (x \right ) \left (u \left (x \right )+2\right )}{2}+\frac {c_{1}}{2 x^{2}}-c_{3} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y \left (\frac {y}{x}+2\right )}{2 x}+\frac {c_{1}}{2 x^{2}}-c_{3} = 0\\ \frac {-2 c_{3} x^{2}+y^{2}+2 x y+c_{1}}{2 x^{2}} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {-2 c_{3} x^{2}+y^{2}+2 x y+c_{1}}{2 x^{2}} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {-2 c_{3} x^{2}+y^{2}+2 x y+c_{1}}{2 x^{2}} = 0 \] Verified OK.
Writing the ode as \[ \left (x y+x^{2}\right ) y^{\prime \prime }+\left (\left (y^{\prime }+1\right ) x -y\right ) y^{\prime }-y = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (\left (x y+x^{2}\right ) y^{\prime \prime }+\left (\left (y^{\prime }+1\right ) x -y\right ) y^{\prime }-y\right )d x &= 0 \\ y^{\prime } x y+y^{\prime } x^{2}-x y-y^{2} = c_{1} \end {align*}
Which is now solved for \(y\). Using the change of variables \(y = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2} u \left (x \right )+\left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x^{2}-x^{2} u \left (x \right )-u \left (x \right )^{2} x^{2} = c_{1} \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {c_{1}}{x^{3} \left (u +1\right )} \end {align*}
Where \(f(x)=\frac {c_{1}}{x^{3}}\) and \(g(u)=\frac {1}{u +1}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{u +1}} \,du &= \frac {c_{1}}{x^{3}} \,d x \\ \int { \frac {1}{\frac {1}{u +1}} \,du} &= \int {\frac {c_{1}}{x^{3}} \,d x} \\ \frac {u \left (u +2\right )}{2}&=-\frac {c_{1}}{2 x^{2}}+c_{3} \\ \end{align*} The solution is \[ \frac {u \left (x \right ) \left (u \left (x \right )+2\right )}{2}+\frac {c_{1}}{2 x^{2}}-c_{3} = 0 \] Replacing \(u(x)\) in the above solution by \(\frac {y}{x}\) results in the solution for \(y\) in implicit form \begin {align*} \frac {y \left (\frac {y}{x}+2\right )}{2 x}+\frac {c_{1}}{2 x^{2}}-c_{3} = 0\\ \frac {-2 c_{3} x^{2}+y^{2}+2 x y+c_{1}}{2 x^{2}} = 0 \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \frac {-2 c_{3} x^{2}+y^{2}+2 x y+c_{1}}{2 x^{2}} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {-2 c_{3} x^{2}+y^{2}+2 x y+c_{1}}{2 x^{2}} = 0 \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN differential order: 2; trying a linearization to 3rd order trying 2nd order ODE linearizable_by_differentiation trying 2nd order, 2 integrating factors of the form mu(x,y) trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type <- LODE of Euler type successful <- 2nd order, 2 integrating factors of the form mu(x,y) successful`
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 49
dsolve(x*(x+y(x))*diff(diff(y(x),x),x)+x*diff(y(x),x)^2+(x-y(x))*diff(y(x),x)-y(x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= -x \\ y \left (x \right ) &= -x -\sqrt {\left (-c_{2} +1\right ) x^{2}+c_{1}} \\ y \left (x \right ) &= -x +\sqrt {\left (-c_{2} +1\right ) x^{2}+c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.141 (sec). Leaf size: 53
DSolve[-y[x] + (x - y[x])*y'[x] + x*y'[x]^2 + x*(x + y[x])*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -x-\sqrt {(1+2 c_2) x^2+c_1} \\ y(x)\to -x+\sqrt {(1+2 c_2) x^2+c_1} \\ \end{align*}