Internal problem ID [9968]
Internal file name [OUTPUT/8915_Monday_June_06_2022_05_50_48_AM_80579890/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1646 (6.56).
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime }+a y \left ({y^{\prime }}^{2}+1\right )^{2}=0} \]
This is missing independent variable second order ode. Solved by reduction of order by using substitution which makes the dependent variable \(y\) an independent variable. Using \begin {align*} y' &= p(y) \end {align*}
Then \begin {align*} y'' &= \frac {dp}{dx}\\ &= \frac {dy}{dx} \frac {dp}{dy}\\ &= p \frac {dp}{dy} \end {align*}
Hence the ode becomes \begin {align*} p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (a y p \left (y \right )^{3}+2 a y p \left (y \right )\right ) p \left (y \right )+a y = 0 \end {align*}
Which is now solved as first order ode for \(p(y)\). In canonical form the ODE is \begin {align*} p' &= F(y,p)\\ &= f( y) g(p)\\ &= -\frac {a y \left (p^{2}+1\right )^{2}}{p} \end {align*}
Where \(f(y)=-a y\) and \(g(p)=\frac {\left (p^{2}+1\right )^{2}}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {\left (p^{2}+1\right )^{2}}{p}} \,dp &= -a y \,d y \\ \int { \frac {1}{\frac {\left (p^{2}+1\right )^{2}}{p}} \,dp} &= \int {-a y \,d y} \\ -\frac {1}{2 \left (p^{2}+1\right )}&=-\frac {a \,y^{2}}{2}+c_{1} \\ \end{align*} The solution is \[ -\frac {1}{2 \left (p \left (y \right )^{2}+1\right )}+\frac {a \,y^{2}}{2}-c_{1} = 0 \] 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 {1}{2 \left ({y^{\prime }}^{2}+1\right )}+\frac {a y^{2}}{2}-c_{1} = 0 \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 {\sqrt {-\left (a y^{2}-2 c_{1} \right ) \left (a y^{2}-2 c_{1} -1\right )}}{a y^{2}-2 c_{1}} \tag {1} \\ y^{\prime }&=\frac {\sqrt {-\left (a y^{2}-2 c_{1} \right ) \left (a y^{2}-2 c_{1} -1\right )}}{a y^{2}-2 c_{1}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int -\frac {a \,y^{2}-2 c_{1}}{\sqrt {-\left (a \,y^{2}-2 c_{1} \right ) \left (a \,y^{2}-2 c_{1} -1\right )}}d y &= \int {dx}\\ \int _{}^{y}-\frac {\textit {\_a}^{2} a -2 c_{1}}{\sqrt {-\left (\textit {\_a}^{2} a -2 c_{1} \right ) \left (\textit {\_a}^{2} a -2 c_{1} -1\right )}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int \frac {a \,y^{2}-2 c_{1}}{\sqrt {-\left (a \,y^{2}-2 c_{1} \right ) \left (a \,y^{2}-2 c_{1} -1\right )}}d y &= \int {dx}\\ \int _{}^{y}\frac {\textit {\_a}^{2} a -2 c_{1}}{\sqrt {-\left (\textit {\_a}^{2} a -2 c_{1} \right ) \left (\textit {\_a}^{2} a -2 c_{1} -1\right )}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}-\frac {\textit {\_a}^{2} a -2 c_{1}}{\sqrt {-\left (\textit {\_a}^{2} a -2 c_{1} \right ) \left (\textit {\_a}^{2} a -2 c_{1} -1\right )}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}\frac {\textit {\_a}^{2} a -2 c_{1}}{\sqrt {-\left (\textit {\_a}^{2} a -2 c_{1} \right ) \left (\textit {\_a}^{2} a -2 c_{1} -1\right )}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}-\frac {\textit {\_a}^{2} a -2 c_{1}}{\sqrt {-\left (\textit {\_a}^{2} a -2 c_{1} \right ) \left (\textit {\_a}^{2} a -2 c_{1} -1\right )}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}\frac {\textit {\_a}^{2} a -2 c_{1}}{\sqrt {-\left (\textit {\_a}^{2} a -2 c_{1} \right ) \left (\textit {\_a}^{2} a -2 c_{1} -1\right )}}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\left (a y {y^{\prime }}^{3}+2 a y y^{\prime }\right ) y^{\prime }+a y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+\left (a y u \left (y \right )^{3}+2 a y u \left (y \right )\right ) u \left (y \right )+a y =0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {\left (a y u \left (y \right )^{3}+2 a y u \left (y \right )\right ) u \left (y \right )+a y}{u \left (y \right )} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{\left (u \left (y \right )^{2}+1\right )^{2}}=-a y \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\left (\frac {d}{d y}u \left (y \right )\right ) u \left (y \right )}{\left (u \left (y \right )^{2}+1\right )^{2}}d y =\int -a y d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{2 \left (u \left (y \right )^{2}+1\right )}=-\frac {a \,y^{2}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\frac {\sqrt {-\left (-a \,y^{2}+2 c_{1} \right ) \left (-a \,y^{2}+2 c_{1} +1\right )}}{-a \,y^{2}+2 c_{1}}, u \left (y \right )=-\frac {\sqrt {-\left (-a \,y^{2}+2 c_{1} \right ) \left (-a \,y^{2}+2 c_{1} +1\right )}}{-a \,y^{2}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\frac {\sqrt {-\left (-a \,y^{2}+2 c_{1} \right ) \left (-a \,y^{2}+2 c_{1} +1\right )}}{-a \,y^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\frac {\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}{-a y^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}{-a y^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } \left (-a y^{2}+2 c_{1} \right )}{\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } \left (-a y^{2}+2 c_{1} \right )}{\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {a \left (-4 c_{1}^{2}-2 c_{1} \right ) \sqrt {1-\frac {a y^{2}}{2 c_{1} +1}}\, \sqrt {4-\frac {2 a y^{2}}{c_{1}}}\, \left (\mathit {EllipticF}\left (y \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {-4+\frac {2 \left (4 c_{1} a +a \right )}{a c_{1}}}}{2}\right )-\mathit {EllipticE}\left (y \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {-4+\frac {2 \left (4 c_{1} a +a \right )}{a c_{1}}}}{2}\right )\right )}{\sqrt {\frac {a}{2 c_{1} +1}}\, \sqrt {-y^{4} a^{2}+4 y^{2} a c_{1} +a y^{2}-4 c_{1}^{2}-2 c_{1}}\, \left (4 c_{1} a +2 a \right )}+\frac {c_{1} \sqrt {1-\frac {a y^{2}}{2 c_{1} +1}}\, \sqrt {4-\frac {2 a y^{2}}{c_{1}}}\, \mathit {EllipticF}\left (y \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {-4+\frac {2 \left (4 c_{1} a +a \right )}{a c_{1}}}}{2}\right )}{\sqrt {\frac {a}{2 c_{1} +1}}\, \sqrt {-y^{4} a^{2}+4 y^{2} a c_{1} +a y^{2}-4 c_{1}^{2}-2 c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\mathit {RootOf}\left (2 \mathit {EllipticE}\left (\textit {\_Z} \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {2}\, \sqrt {\frac {2 c_{1} +1}{c_{1}}}}{2}\right ) c_{1} -\sqrt {-2 c_{1} a}\, c_{2} -\sqrt {-2 c_{1} a}\, x \right ), \mathit {RootOf}\left (2 \mathit {EllipticE}\left (\textit {\_Z} \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {2}\, \sqrt {\frac {2 c_{1} +1}{c_{1}}}}{2}\right ) c_{1} +\sqrt {-2 c_{1} a}\, c_{2} +\sqrt {-2 c_{1} a}\, x \right )\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\frac {\sqrt {-\left (-a \,y^{2}+2 c_{1} \right ) \left (-a \,y^{2}+2 c_{1} +1\right )}}{-a \,y^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\frac {\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}{-a y^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}{-a y^{2}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime } \left (-a y^{2}+2 c_{1} \right )}{\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime } \left (-a y^{2}+2 c_{1} \right )}{\sqrt {-\left (-a y^{2}+2 c_{1} \right ) \left (-a y^{2}+2 c_{1} +1\right )}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {a \left (-4 c_{1}^{2}-2 c_{1} \right ) \sqrt {1-\frac {a y^{2}}{2 c_{1} +1}}\, \sqrt {4-\frac {2 a y^{2}}{c_{1}}}\, \left (\mathit {EllipticF}\left (y \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {-4+\frac {2 \left (4 c_{1} a +a \right )}{a c_{1}}}}{2}\right )-\mathit {EllipticE}\left (y \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {-4+\frac {2 \left (4 c_{1} a +a \right )}{a c_{1}}}}{2}\right )\right )}{\sqrt {\frac {a}{2 c_{1} +1}}\, \sqrt {-y^{4} a^{2}+4 y^{2} a c_{1} +a y^{2}-4 c_{1}^{2}-2 c_{1}}\, \left (4 c_{1} a +2 a \right )}+\frac {c_{1} \sqrt {1-\frac {a y^{2}}{2 c_{1} +1}}\, \sqrt {4-\frac {2 a y^{2}}{c_{1}}}\, \mathit {EllipticF}\left (y \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {-4+\frac {2 \left (4 c_{1} a +a \right )}{a c_{1}}}}{2}\right )}{\sqrt {\frac {a}{2 c_{1} +1}}\, \sqrt {-y^{4} a^{2}+4 y^{2} a c_{1} +a y^{2}-4 c_{1}^{2}-2 c_{1}}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\mathit {RootOf}\left (2 \mathit {EllipticE}\left (\textit {\_Z} \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {2}\, \sqrt {\frac {2 c_{1} +1}{c_{1}}}}{2}\right ) c_{1} -\sqrt {-2 c_{1} a}\, c_{2} +\sqrt {-2 c_{1} a}\, x \right ), \mathit {RootOf}\left (2 \mathit {EllipticE}\left (\textit {\_Z} \sqrt {\frac {a}{2 c_{1} +1}}, \frac {\sqrt {2}\, \sqrt {\frac {2 c_{1} +1}{c_{1}}}}{2}\right ) c_{1} +\sqrt {-2 c_{1} a}\, c_{2} -\sqrt {-2 c_{1} a}\, x \right )\right \} \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 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 `, `-> Computing symmetries using: way = exp_sym -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+a*_a*(_b(_a)^2+1)^2 = 0, _b(_a)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable <- separable successful <- differential order: 2; canonical coordinates successful <- differential order 2; missing variables successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 96
dsolve(diff(diff(y(x),x),x)+a*y(x)*(diff(y(x),x)^2+1)^2=0,y(x), singsol=all)
\begin{align*} a \left (\int _{}^{y \left (x \right )}\frac {\textit {\_a}^{2}+2 c_{1}}{\sqrt {-\left (-1+a \left (\textit {\_a}^{2}+2 c_{1} \right )\right ) \left (\textit {\_a}^{2}+2 c_{1} \right ) a}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ -a \left (\int _{}^{y \left (x \right )}\frac {\textit {\_a}^{2}+2 c_{1}}{\sqrt {-\left (-1+a \left (\textit {\_a}^{2}+2 c_{1} \right )\right ) \left (\textit {\_a}^{2}+2 c_{1} \right ) a}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 22.617 (sec). Leaf size: 816
DSolve[a*y[x]*(1 + y'[x]^2)^2 + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {\frac {\text {$\#$1}^2 (-a)+1+2 c_1}{1+2 c_1}} \sqrt {2 \text {$\#$1}^2 a-4 c_1} E\left (\arcsin \left (\sqrt {\frac {a}{2 c_1+1}} \text {$\#$1}\right )|1+\frac {1}{2 c_1}\right )}{\sqrt {\frac {a}{1+2 c_1}} \sqrt {\text {$\#$1}^2 (-a)+1+2 c_1} \sqrt {2-\frac {\text {$\#$1}^2 a}{c_1}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {\frac {\text {$\#$1}^2 (-a)+1+2 c_1}{1+2 c_1}} \sqrt {2 \text {$\#$1}^2 a-4 c_1} E\left (\arcsin \left (\sqrt {\frac {a}{2 c_1+1}} \text {$\#$1}\right )|1+\frac {1}{2 c_1}\right )}{\sqrt {\frac {a}{1+2 c_1}} \sqrt {\text {$\#$1}^2 (-a)+1+2 c_1} \sqrt {2-\frac {\text {$\#$1}^2 a}{c_1}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {\frac {\text {$\#$1}^2 (-a)+1+2 (-1) c_1}{1+2 (-1) c_1}} \sqrt {2 \text {$\#$1}^2 a-4 (-c_1)} E\left (\arcsin \left (\sqrt {\frac {a}{2 (-1) c_1+1}} \text {$\#$1}\right )|1+\frac {1}{2 (-c_1)}\right )}{\sqrt {\frac {a}{1+2 (-1) c_1}} \sqrt {\text {$\#$1}^2 (-a)+1+2 (-1) c_1} \sqrt {2--\frac {\text {$\#$1}^2 a}{c_1}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {\frac {\text {$\#$1}^2 (-a)+1+2 (-1) c_1}{1+2 (-1) c_1}} \sqrt {2 \text {$\#$1}^2 a-4 (-c_1)} E\left (\arcsin \left (\sqrt {\frac {a}{2 (-1) c_1+1}} \text {$\#$1}\right )|1+\frac {1}{2 (-c_1)}\right )}{\sqrt {\frac {a}{1+2 (-1) c_1}} \sqrt {\text {$\#$1}^2 (-a)+1+2 (-1) c_1} \sqrt {2--\frac {\text {$\#$1}^2 a}{c_1}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt {\frac {\text {$\#$1}^2 (-a)+1+2 c_1}{1+2 c_1}} \sqrt {2 \text {$\#$1}^2 a-4 c_1} E\left (\arcsin \left (\sqrt {\frac {a}{2 c_1+1}} \text {$\#$1}\right )|1+\frac {1}{2 c_1}\right )}{\sqrt {\frac {a}{1+2 c_1}} \sqrt {\text {$\#$1}^2 (-a)+1+2 c_1} \sqrt {2-\frac {\text {$\#$1}^2 a}{c_1}}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\sqrt {\frac {\text {$\#$1}^2 (-a)+1+2 c_1}{1+2 c_1}} \sqrt {2 \text {$\#$1}^2 a-4 c_1} E\left (\arcsin \left (\sqrt {\frac {a}{2 c_1+1}} \text {$\#$1}\right )|1+\frac {1}{2 c_1}\right )}{\sqrt {\frac {a}{1+2 c_1}} \sqrt {\text {$\#$1}^2 (-a)+1+2 c_1} \sqrt {2-\frac {\text {$\#$1}^2 a}{c_1}}}\&\right ][x+c_2] \\ \end{align*}