Internal problem ID [10078]
Internal file name [OUTPUT/9025_Monday_June_06_2022_06_14_26_AM_69378563/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1757 (book 6.166).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_nonlinear_solved_by_mainardi_lioville_method"
Maple gives the following as the ode type
[_Liouville, [_2nd_order, _with_linear_symmetries], [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {a y y^{\prime \prime }+{y^{\prime }}^{2} b -\frac {y y^{\prime }}{\sqrt {c^{2}+x^{2}}}=0} \]
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}{a \sqrt {c^{2}+x^{2}}}\\ g(y) &= \frac {b}{a 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 \(g=\frac {b}{a y}\) and \(f=-\frac {1}{a \sqrt {c^{2}+x^{2}}}\), then \begin {align*} \int -g d y &= \int -\frac {b}{a y}d y\\ &= -\frac {b \ln \left (y\right )}{a}\\ \int -f d x &= \int \frac {1}{a \sqrt {c^{2}+x^{2}}}d x\\ &= \frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a} \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = c_{2} {\mathrm e}^{-\frac {b \ln \left (y\right )}{a}} {\mathrm e}^{\frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a}} \] 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)\\ &= c_{2} {\mathrm e}^{\frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a}} y^{-\frac {b}{a}} \end {align*}
Where \(f(x)=c_{2} {\mathrm e}^{\frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a}}\) and \(g(y)=y^{-\frac {b}{a}}\). Integrating both sides gives \begin{align*} \frac {1}{y^{-\frac {b}{a}}} \,dy &= c_{2} {\mathrm e}^{\frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a}} \,d x \\ \int { \frac {1}{y^{-\frac {b}{a}}} \,dy} &= \int {c_{2} {\mathrm e}^{\frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a}} \,d x} \\ \frac {y a \,y^{\frac {b}{a}}}{a +b}&=\int c_{2} {\mathrm e}^{\frac {\ln \left (x +\sqrt {c^{2}+x^{2}}\right )}{a}}d x +c_{3} \\ \end{align*} Which simplifies to \[ \frac {y a y^{\frac {b}{a}}}{a +b}-\left (\int c_{2} \left (x +\sqrt {c^{2}+x^{2}}\right )^{\frac {1}{a}}d x \right )-c_{3} = 0 \] The solution is \[ \frac {y a y^{\frac {b}{a}}}{a +b}-\left (\int c_{2} \left (x +\sqrt {c^{2}+x^{2}}\right )^{\frac {1}{a}}d x \right )-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y a y^{\frac {b}{a}}}{a +b}-\left (\int c_{2} \left (x +\sqrt {c^{2}+x^{2}}\right )^{\frac {1}{a}}d x \right )-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y a y^{\frac {b}{a}}}{a +b}-\left (\int c_{2} \left (x +\sqrt {c^{2}+x^{2}}\right )^{\frac {1}{a}}d x \right )-c_{3} = 0 \] 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: 0.062 (sec). Leaf size: 82
dsolve(a*y(x)*diff(diff(y(x),x),x)+b*diff(y(x),x)^2-y(x)*diff(y(x),x)/(c^2+x^2)^(1/2)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= {\left (\frac {a \left (a +1\right )}{\left (a +b \right ) \left (c_{1} 2^{\frac {1}{a}} a \,x^{\frac {a +1}{a}} \operatorname {hypergeom}\left (\left [-\frac {1}{2 a}, -\frac {a +1}{2 a}\right ], \left [\frac {a -1}{a}\right ], -\frac {c^{2}}{x^{2}}\right )+c_{2} a +c_{2} \right )}\right )}^{-\frac {a}{a +b}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 66.528 (sec). Leaf size: 143
DSolve[-((y[x]*y'[x])/Sqrt[c^2 + x^2]) + b*y'[x]^2 + a*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \exp \left (\int _1^x\frac {\left (1-\frac {K[2]}{\sqrt {c^2+K[2]^2}}\right )^{\left .-\frac {1}{2}\right /a} \left (\frac {K[2]}{\sqrt {c^2+K[2]^2}}+1\right )^{\left .\frac {1}{2}\right /a}}{c_1-\int _1^{K[2]}-\frac {(a+b) \left (1-\frac {K[1]}{\sqrt {c^2+K[1]^2}}\right )^{\left .-\frac {1}{2}\right /a} \left (\frac {K[1]}{\sqrt {c^2+K[1]^2}}+1\right )^{\left .\frac {1}{2}\right /a}}{a}dK[1]}dK[2]\right ) \]