Internal problem ID [10149]
Internal file name [OUTPUT/9096_Monday_June_06_2022_06_34_26_AM_96531773/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1828 (book 6.237).
ODE order: 2.
ODE degree: 2.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {a^{2} {y^{\prime \prime }}^{2}-2 a x y^{\prime \prime }+y^{\prime }=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} \left (a^{2} p^{\prime }\left (x \right )-2 a x \right ) p^{\prime }\left (x \right )+p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Let \(p=p^{\prime }\left (x \right )\) the ode becomes \begin {align*} \left (a^{2} p -2 a x \right ) p +p = 0 \end {align*}
Solving for \(p \left (x \right )\) from the above results in \begin {align*} p \left (x \right ) &= -a^{2} p^{2}+2 a p x\tag {1A} \end {align*}
This has the form \begin {align*} p=xf(p)+g(p)\tag {*} \end {align*}
Where \(f,g\) are functions of \(p=p'(x)\). The above ode is dAlembert ode which is now solved. Taking derivative of (*) w.r.t. \(x\) gives \begin {align*} p &= f+(x f'+g') \frac {dp}{dx}\\ p-f &= (x f'+g') \frac {dp}{dx}\tag {2} \end {align*}
Comparing the form \(p \left (x \right )=x f + g\) to (1A) shows that \begin {align*} f &= 2 a p\\ g &= -a^{2} p^{2} \end {align*}
Hence (2) becomes \begin {align*} -2 a p +p = \left (-2 a^{2} p +2 a x \right ) p^{\prime }\left (x \right )\tag {2A} \end {align*}
The singular solution is found by setting \(\frac {dp}{dx}=0\) in the above which gives \begin {align*} -2 a p +p = 0 \end {align*}
Solving for \(p\) from the above gives \begin {align*} p&=0 \end {align*}
Substituting these in (1A) gives \begin {align*} p \left (x \right )&=0 \end {align*}
The general solution is found when \( \frac { \mathop {\mathrm {d}p}}{\mathop {\mathrm {d}x}}\neq 0\). From eq. (2A). This results in \begin {align*} p^{\prime }\left (x \right ) = \frac {-2 a p \left (x \right )+p \left (x \right )}{-2 a^{2} p \left (x \right )+2 a x}\tag {3} \end {align*}
This ODE is now solved for \(p \left (x \right )\).
Inverting the above ode gives \begin {align*} \frac {d}{d p}x \left (p \right ) = \frac {-2 a^{2} p +2 a x \left (p \right )}{-2 a p +p}\tag {4} \end {align*}
This ODE is now solved for \(x \left (p \right )\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} \frac {d}{d p}x \left (p \right ) + p(p)x \left (p \right ) &= q(p) \end {align*}
Where here \begin {align*} p(p) &=\frac {2 a}{\left (2 a -1\right ) p}\\ q(p) &=\frac {2 a^{2}}{2 a -1} \end {align*}
Hence the ode is \begin {align*} \frac {d}{d p}x \left (p \right )+\frac {2 a x \left (p \right )}{\left (2 a -1\right ) p} = \frac {2 a^{2}}{2 a -1} \end {align*}
The integrating factor \(\mu \) is \begin{align*} \mu &= {\mathrm e}^{\int \frac {2 a}{\left (2 a -1\right ) p}d p} \\ &= {\mathrm e}^{\frac {2 a \ln \left (p \right )}{2 a -1}} \\ \end{align*} Which simplifies to \[ \mu = p^{\frac {2 a}{2 a -1}} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}}\left ( \mu x\right ) &= \left (\mu \right ) \left (\frac {2 a^{2}}{2 a -1}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}p}} \left (p^{\frac {2 a}{2 a -1}} x\right ) &= \left (p^{\frac {2 a}{2 a -1}}\right ) \left (\frac {2 a^{2}}{2 a -1}\right )\\ \mathrm {d} \left (p^{\frac {2 a}{2 a -1}} x\right ) &= \left (\frac {2 a^{2} p^{\frac {2 a}{2 a -1}}}{2 a -1}\right )\, \mathrm {d} p \end {align*}
Integrating gives \begin {align*} p^{\frac {2 a}{2 a -1}} x &= \int {\frac {2 a^{2} p^{\frac {2 a}{2 a -1}}}{2 a -1}\,\mathrm {d} p}\\ p^{\frac {2 a}{2 a -1}} x &= \frac {2 p^{1+\frac {2 a}{2 a -1}} a^{2}}{4 a -1} + c_{1} \end {align*}
Dividing both sides by the integrating factor \(\mu =p^{\frac {2 a}{2 a -1}}\) results in \begin {align*} x \left (p \right ) &= \frac {2 p^{-\frac {2 a}{2 a -1}} p^{1+\frac {2 a}{2 a -1}} a^{2}}{4 a -1}+c_{1} p^{-\frac {2 a}{2 a -1}} \end {align*}
which simplifies to \begin {align*} x \left (p \right ) &= \frac {2 a^{2} p}{4 a -1}+c_{1} p^{-\frac {2 a}{2 a -1}} \end {align*}
Now we need to eliminate \(p\) between the above and (1A). One way to do this is by solving (1) for \(p\). This results in \begin {align*} p&=\frac {x +\sqrt {x^{2}-p \left (x \right )}}{a}\\ p&=\frac {x -\sqrt {x^{2}-p \left (x \right )}}{a} \end {align*}
Substituting the above in the solution for \(x\) found above gives \begin{align*} x&=\frac {c_{1} \left (4 a -1\right ) {\left (\frac {x +\sqrt {x^{2}-p \left (x \right )}}{a}\right )}^{-\frac {2 a}{2 a -1}}+2 a \left (x +\sqrt {x^{2}-p \left (x \right )}\right )}{4 a -1} \\ x&=\frac {c_{1} \left (4 a -1\right ) {\left (\frac {x -\sqrt {x^{2}-p \left (x \right )}}{a}\right )}^{-\frac {2 a}{2 a -1}}+2 a \left (x -\sqrt {x^{2}-p \left (x \right )}\right )}{4 a -1} \\ \end{align*} For solution (1) found earlier, since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = 0 \end {align*}
Integrating both sides gives \begin {align*} y &= \int { 0\,\mathop {\mathrm {d}x}}\\ &= c_{2} \end {align*}
For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} x = \frac {c_{1} \left (4 a -1\right ) {\left (\frac {x +\sqrt {x^{2}-y^{\prime }}}{a}\right )}^{-\frac {2 a}{2 a -1}}+2 a \left (x +\sqrt {x^{2}-y^{\prime }}\right )}{4 a -1} \end {align*}
Integrating both sides gives \begin {align*} y = \int \left (-\operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right )^{2} a^{2}+2 \operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right ) a x \right )d x +c_{3} \end {align*}
For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} x = \frac {c_{1} \left (4 a -1\right ) {\left (\frac {x -\sqrt {x^{2}-y^{\prime }}}{a}\right )}^{-\frac {2 a}{2 a -1}}+2 a \left (x -\sqrt {x^{2}-y^{\prime }}\right )}{4 a -1} \end {align*}
Integrating both sides gives \begin {align*} y = \int \left (-\operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right )^{2} a^{2}+2 \operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right ) a x \right )d x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{2} \\ \tag{2} y &= \int \left (-\operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right )^{2} a^{2}+2 \operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right ) a x \right )d x +c_{3} \\ \tag{3} y &= \int \left (-\operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right )^{2} a^{2}+2 \operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right ) a x \right )d x +c_{4} \\ \end{align*}
Verification of solutions
\[ y = c_{2} \] Verified OK.
\[ y = \int \left (-\operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right )^{2} a^{2}+2 \operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right ) a x \right )d x +c_{3} \] Verified OK.
\[ y = \int \left (-\operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right )^{2} a^{2}+2 \operatorname {RootOf}\left (2 \textit {\_Z} \,a^{2} \textit {\_Z}^{\frac {2 a}{2 a -1}}-4 a x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+x \,\textit {\_Z}^{\frac {2 a}{2 a -1}}+4 c_{1} a -c_{1} \right ) a x \right )d x +c_{4} \] Verified OK.
Maple trace
`Methods for second order ODEs: Successful isolation of d^2y/dx^2: 2 solutions were found. Trying to solve each resulting ODE. *** Sublevel 2 *** 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 differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(-_a+(_a^2-_b(_a))^(1/2))/a, _b(_a), HINT = [[_a, 2*_b]]` *** Sublevel symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 2*_b]
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 1364
dsolve(a^2*diff(diff(y(x),x),x)^2-2*a*x*diff(diff(y(x),x),x)+diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \int \operatorname {RootOf}\left (8 \textit {\_Z}^{1-2 a} a^{3} \sqrt {x^{2}-\textit {\_Z}}\, \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a}-8 \sqrt {x^{2}-\textit {\_Z}}\, \textit {\_Z}^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} a^{2} x^{2}+8 x \,\textit {\_Z}^{1-2 a} a^{3} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a}-8 \textit {\_Z}^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} a^{2} x^{3}+2 \sqrt {x^{2}-\textit {\_Z}}\, \textit {\_Z}^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} a \,x^{2}-4 x \,\textit {\_Z}^{1-2 a} a^{2} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a}+6 \textit {\_Z}^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} a \,x^{3}-\textit {\_Z}^{-2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} x^{3}-2 \sqrt {x^{2}-\textit {\_Z}}\, c_{1} a +2 c_{1} a x -c_{1} x \right )d x +c_{2} \\ y \left (x \right ) &= \int \operatorname {RootOf}\left (2 \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{2 a} \sqrt {x^{2}-\textit {\_Z}}\, \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} \textit {\_Z}^{2 a} a +2 \left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} \textit {\_Z}^{2 a} a x -\left (4 \textit {\_Z} \,a^{2}-4 a \,x^{2}+x^{2}\right )^{2 a} \left (x +\sqrt {x^{2}-\textit {\_Z}}\right )^{2 a} \left (-x +\sqrt {x^{2}-\textit {\_Z}}\right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}+2 a x -x \right )^{-2 a} \left (2 a \sqrt {x^{2}-\textit {\_Z}}-2 a x +x \right )^{2 a} \textit {\_Z}^{2 a} x -8 \sqrt {x^{2}-\textit {\_Z}}\, c_{1} \textit {\_Z} \,a^{3}+8 \sqrt {x^{2}-\textit {\_Z}}\, c_{1} a^{2} x^{2}+8 c_{1} \textit {\_Z} \,a^{3} x -8 c_{1} a^{2} x^{3}-2 \sqrt {x^{2}-\textit {\_Z}}\, c_{1} a \,x^{2}-4 c_{1} \textit {\_Z} \,a^{2} x +6 c_{1} a \,x^{3}-c_{1} x^{3}\right )d x +c_{2} \\ \end{align*}
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x] - 2*a*x*y''[x] + a^2*y''[x]^2 == 0,y[x],x,IncludeSingularSolutions -> True]
Not solved