Internal problem ID [9993]
Internal file name [OUTPUT/8940_Monday_June_06_2022_06_00_42_AM_47463325/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1672 (book 6.81).
ODE order: 2.
ODE degree: 1.
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], [_2nd_order, _reducible, _mu_y_y1]]
\[ \boxed {2 x y^{\prime \prime }+{y^{\prime }}^{3}+y^{\prime }=0} \]
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*} 2 x p^{\prime }\left (x \right )+\left (p \left (x \right )^{2}+1\right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= -\frac {\left (p^{2}+1\right ) p}{2 x} \end {align*}
Where \(f(x)=-\frac {1}{2 x}\) and \(g(p)=p \left (p^{2}+1\right )\). Integrating both sides gives \begin{align*} \frac {1}{p \left (p^{2}+1\right )} \,dp &= -\frac {1}{2 x} \,d x \\ \int { \frac {1}{p \left (p^{2}+1\right )} \,dp} &= \int {-\frac {1}{2 x} \,d x} \\ -\frac {\ln \left (p^{2}+1\right )}{2}+\ln \left (p \right )&=-\frac {\ln \left (x \right )}{2}+c_{1} \\ \end{align*} Raising both side to exponential gives \begin {align*} {\mathrm e}^{-\frac {\ln \left (p^{2}+1\right )}{2}+\ln \left (p \right )} &= {\mathrm e}^{-\frac {\ln \left (x \right )}{2}+c_{1}} \end {align*}
Which simplifies to \begin {align*} \frac {p}{\sqrt {p^{2}+1}} &= \frac {c_{2}}{\sqrt {x}} \end {align*}
The solution is \[ \frac {p \left (x \right )}{\sqrt {p \left (x \right )^{2}+1}} = \frac {c_{2}}{\sqrt {x}} \] For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is \begin {align*} \frac {y^{\prime }}{\sqrt {{y^{\prime }}^{2}+1}} = \frac {c_{2}}{\sqrt {x}} \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=-\frac {c_{2}}{\sqrt {-c_{2}^{2}+x}} \tag {1} \\ y^{\prime }&=\frac {c_{2}}{\sqrt {-c_{2}^{2}+x}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} y &= \int { -\frac {c_{2}}{\sqrt {-c_{2}^{2}+x}}\,\mathop {\mathrm {d}x}}\\ &= -2 \sqrt {-c_{2}^{2}+x}\, c_{2} +c_{3} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} y &= \int { \frac {c_{2}}{\sqrt {-c_{2}^{2}+x}}\,\mathop {\mathrm {d}x}}\\ &= 2 \sqrt {-c_{2}^{2}+x}\, c_{2} +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -2 \sqrt {-c_{2}^{2}+x}\, c_{2} +c_{3} \\ \tag{2} y &= 2 \sqrt {-c_{2}^{2}+x}\, c_{2} +c_{4} \\ \end{align*}
Verification of solutions
\[ y = -2 \sqrt {-c_{2}^{2}+x}\, c_{2} +c_{3} \] Verified OK.
\[ y = 2 \sqrt {-c_{2}^{2}+x}\, c_{2} +c_{4} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x \left (\frac {d}{d x}y^{\prime }\right )+\left ({y^{\prime }}^{2}+1\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime }\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & 2 x u^{\prime }\left (x \right )+\left (u \left (x \right )^{2}+1\right ) u \left (x \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & u^{\prime }\left (x \right )=-\frac {\left (u \left (x \right )^{2}+1\right ) u \left (x \right )}{2 x} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {u^{\prime }\left (x \right )}{\left (u \left (x \right )^{2}+1\right ) u \left (x \right )}=-\frac {1}{2 x} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {u^{\prime }\left (x \right )}{\left (u \left (x \right )^{2}+1\right ) u \left (x \right )}d x =\int -\frac {1}{2 x}d x +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {\ln \left (u \left (x \right )^{2}+1\right )}{2}+\ln \left (u \left (x \right )\right )=-\frac {\ln \left (x \right )}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (x \right ) \\ {} & {} & \left \{u \left (x \right )=\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}}, u \left (x \right )=-\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int \frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=-2 \sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}+c_{2} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (x \right ) \\ {} & {} & u \left (x \right )=-\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}} \\ \bullet & {} & \textrm {Make substitution}\hspace {3pt} u =y^{\prime } \\ {} & {} & y^{\prime }=-\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}} \\ \bullet & {} & \textrm {Integrate both sides to solve for}\hspace {3pt} y \\ {} & {} & \int y^{\prime }d x =\int -\frac {\sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}}{-x +{\mathrm e}^{2 c_{1}}}d x +c_{2} \\ \bullet & {} & \textrm {Compute integrals}\hspace {3pt} \\ {} & {} & y=2 \sqrt {-\left (-x +{\mathrm e}^{2 c_{1}}\right ) {\mathrm e}^{2 c_{1}}}+c_{2} \end {array} \]
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 a change of variables {x -> y(x), y(x) -> x} 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) = -(1/2)*_b(_a)*(_b(_a)^2+1)/_a, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 41
dsolve(2*x*diff(diff(y(x),x),x)+diff(y(x),x)^3+diff(y(x),x)=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= \frac {c_{2} c_{1} +2 \sqrt {c_{1} x -1}}{c_{1}} \\ y \left (x \right ) &= \frac {c_{2} c_{1} -2 \sqrt {c_{1} x -1}}{c_{1}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 0.896 (sec). Leaf size: 65
DSolve[y'[x] + y'[x]^3 + 2*x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to c_2-2 i e^{c_1} \sqrt {-x+e^{2 c_1}} \\ y(x)\to 2 i e^{c_1} \sqrt {-x+e^{2 c_1}}+c_2 \\ y(x)\to c_2 \\ \end{align*}