Internal problem ID [10081]
Internal file name [OUTPUT/9028_Monday_June_06_2022_06_14_44_AM_81458591/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1760 (book 6.169).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_integrable_as_is", "second_order_nonlinear_solved_by_mainardi_lioville_method"
Maple gives the following as the ode type
[[_2nd_order, _exact, _nonlinear], _Liouville, [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {x y y^{\prime \prime }+x {y^{\prime }}^{2}-y^{\prime } y=0} \]
Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y y^{\prime \prime }+\left (y^{\prime } x -y\right ) y^{\prime }\right )d x &= 0 \\ y^{\prime } x y-y^{2} = c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y^{2}+c_{1}}{x y} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=\frac {y^{2}+c_{1}}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {y^{2}+c_{1}}{y}} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {y^{2}+c_{1}}{y}} \,dy} &= \int {\frac {1}{x} \,d x} \\ \frac {\ln \left (y^{2}+c_{1} \right )}{2}&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {y^{2}+c_{1}} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sqrt {y^{2}+c_{1}} &= c_{3} x \end {align*}
Which simplifies to \[ \sqrt {y^{2}+c_{1}} = c_{3} {\mathrm e}^{c_{2}} x \] The solution is \[ \sqrt {y^{2}+c_{1}} = c_{3} {\mathrm e}^{c_{2}} x \]
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {y^{2}+c_{1}} &= c_{3} {\mathrm e}^{c_{2}} x \\ \end{align*}
Verification of solutions
\[ \sqrt {y^{2}+c_{1}} = c_{3} {\mathrm e}^{c_{2}} x \] Verified OK.
The ode has the Liouville form given by \begin {align*} y^{\prime \prime }+ f(x) y^{\prime } + g(y) {y^{\prime }}^{2} &= 0 \tag {1A} \end {align*}
Where in this problem \begin {align*} f(x) &= -\frac {1}{x}\\ g(y) &= \frac {1}{y} \end {align*}
Dividing through by \(y^{\prime }\) then Eq (1A) becomes \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}+ f + g y^{\prime } &= 0 \tag {2A} \end {align*}
But the first term in Eq (2A) can be written as \begin {align*} \frac {y^{\prime \prime }}{y^{\prime }}&= \frac {d}{dx} \ln \left ( y^{\prime } \right )\tag {3A} \end {align*}
And the last term in Eq (2A) can be written as \begin {align*} g \frac {dy}{dx}&= \left ( \frac {d}{dy} \int g d y\right ) \frac {dy}{dx} \\ &= \frac {d}{dx} \int g d y\tag {4A} \end {align*}
Substituting (3A,4A) back into (2A) gives \begin {align*} \frac {d}{dx} \ln \left ( y^{\prime } \right ) + \frac {d}{dx} \int g d y &= -f \tag {5A} \end {align*}
Integrating the above w.r.t. \(x\) gives \begin {align*} \ln \left ( y^{\prime } \right ) + \int g d y &= - \int f d x + c_{1} \end {align*}
Where \(c_1\) is arbitrary constant. Taking the exponential of the above gives \begin {align*} y^{\prime } &= c_{2} e^{\int -g d y}\, e^{\int -f d x}\tag {6A} \end {align*}
Where \(c_{2}\) is a new arbitrary constant. But since \(g=\frac {1}{y}\) and \(f=-\frac {1}{x}\), then \begin {align*} \int -g d y &= \int -\frac {1}{y}d y\\ &= -\ln \left (y\right )\\ \int -f d x &= \int \frac {1}{x}d x\\ &= \ln \left (x \right ) \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = \frac {c_{2} x}{y} \] Which is now solved as first order separable ode. In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {c_{2} x}{y} \end {align*}
Where \(f(x)=c_{2} x\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= c_{2} x \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {c_{2} x \,d x} \\ \frac {y^{2}}{2}&=\frac {c_{2} x^{2}}{2}+c_{3} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-\frac {c_{2} x^{2}}{2}-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-\frac {c_{2} x^{2}}{2}-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-\frac {c_{2} x^{2}}{2}-c_{3} = 0 \] Verified OK.
Writing the ode as \[ x y y^{\prime \prime }+\left (y^{\prime } x -y\right ) y^{\prime } = 0 \] Integrating both sides of the ODE w.r.t \(x\) gives \begin {align*} \int \left (x y y^{\prime \prime }+\left (y^{\prime } x -y\right ) y^{\prime }\right )d x &= 0 \\ y^{\prime } x y-y^{2} = c_{1} \end {align*}
Which is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {y^{2}+c_{1}}{x y} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(y)=\frac {y^{2}+c_{1}}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {y^{2}+c_{1}}{y}} \,dy &= \frac {1}{x} \,d x \\ \int { \frac {1}{\frac {y^{2}+c_{1}}{y}} \,dy} &= \int {\frac {1}{x} \,d x} \\ \frac {\ln \left (y^{2}+c_{1} \right )}{2}&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {y^{2}+c_{1}} &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \sqrt {y^{2}+c_{1}} &= c_{3} x \end {align*}
Which simplifies to \[ \sqrt {y^{2}+c_{1}} = c_{3} {\mathrm e}^{c_{2}} x \] The solution is \[ \sqrt {y^{2}+c_{1}} = c_{3} {\mathrm e}^{c_{2}} x \]
The solution(s) found are the following \begin{align*} \tag{1} \sqrt {y^{2}+c_{1}} &= c_{3} {\mathrm e}^{c_{2}} x \\ \end{align*}
Verification of solutions
\[ \sqrt {y^{2}+c_{1}} = c_{3} {\mathrm e}^{c_{2}} x \] Verified OK.
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 35
dsolve(x*y(x)*diff(diff(y(x),x),x)+x*diff(y(x),x)^2-y(x)*diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \sqrt {c_{1} x^{2}+2 c_{2}} \\ y \left (x \right ) &= -\sqrt {c_{1} x^{2}+2 c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.39 (sec). Leaf size: 18
DSolve[-(y[x]*y'[x]) + x*y'[x]^2 + x*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \sqrt {x^2+c_1} \]