Internal problem ID [12183]
Internal file name [OUTPUT/10836_Thursday_September_21_2023_05_47_43_AM_53806614/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 30.
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]]
\[ \boxed {y y^{\prime \prime }+{y^{\prime }}^{2}-\frac {y y^{\prime }}{\sqrt {x^{2}+1}}=0} \]
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}{\sqrt {x^{2}+1}}\\ 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 \(f=-\frac {1}{\sqrt {x^{2}+1}}\), then \begin {align*} \int -g d y &= \int -\frac {1}{y}d y\\ &= -\ln \left (y\right )\\ \int -f d x &= \int \frac {1}{\sqrt {x^{2}+1}}d x\\ &= \operatorname {arcsinh}\left (x \right ) \end {align*}
Substituting the above into Eq(6A) gives \[ y^{\prime } = \frac {c_{2} \left (x +\sqrt {x^{2}+1}\right )}{y} \] 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} \left (x +\sqrt {x^{2}+1}\right )}{y} \end {align*}
Where \(f(x)=c_{2} \left (x +\sqrt {x^{2}+1}\right )\) and \(g(y)=\frac {1}{y}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{y}} \,dy &= c_{2} \left (x +\sqrt {x^{2}+1}\right ) \,d x \\ \int { \frac {1}{\frac {1}{y}} \,dy} &= \int {c_{2} \left (x +\sqrt {x^{2}+1}\right ) \,d x} \\ \frac {y^{2}}{2}&=c_{2} \left (\frac {x^{2}}{2}+\frac {x \sqrt {x^{2}+1}}{2}+\frac {\operatorname {arcsinh}\left (x \right )}{2}\right )+c_{3} \\ \end{align*} The solution is \[ \frac {y^{2}}{2}-c_{2} \left (\frac {x^{2}}{2}+\frac {x \sqrt {x^{2}+1}}{2}+\frac {\operatorname {arcsinh}\left (x \right )}{2}\right )-c_{3} = 0 \]
The solution(s) found are the following \begin{align*} \tag{1} \frac {y^{2}}{2}-c_{2} \left (\frac {x^{2}}{2}+\frac {x \sqrt {x^{2}+1}}{2}+\frac {\operatorname {arcsinh}\left (x \right )}{2}\right )-c_{3} &= 0 \\ \end{align*}
Verification of solutions
\[ \frac {y^{2}}{2}-c_{2} \left (\frac {x^{2}}{2}+\frac {x \sqrt {x^{2}+1}}{2}+\frac {\operatorname {arcsinh}\left (x \right )}{2}\right )-c_{3} = 0 \] 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: 0.078 (sec). Leaf size: 63
dsolve(y(x)*diff(y(x),x$2)+diff(y(x),x)^2= y(x)*diff(y(x),x)/sqrt(1+x^2),y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \sqrt {c_{1} x \sqrt {x^{2}+1}+c_{1} x^{2}+c_{1} \operatorname {arcsinh}\left (x \right )+2 c_{2}} \\ y \left (x \right ) &= -\sqrt {c_{1} x \sqrt {x^{2}+1}+c_{1} x^{2}+c_{1} \operatorname {arcsinh}\left (x \right )+2 c_{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.936 (sec). Leaf size: 73
DSolve[y[x]*y''[x]+y'[x]^2== y[x]*y'[x]/Sqrt[1+x^2],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to c_2 \exp \left (\int _1^x\frac {1}{-K[1] c_1+\sqrt {K[1]^2+1} c_1+K[1]+\left (K[1]-\sqrt {K[1]^2+1}\right ) \log \left (\sqrt {K[1]^2+1}-K[1]\right )}dK[1]\right ) \]