Internal problem ID [12205]
Internal file name [OUTPUT/10858_Thursday_September_21_2023_05_48_11_AM_81150646/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 58.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x", "second_order_ode_can_be_made_integrable"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime \prime }-2 y^{3}=0} \] With initial conditions \begin {align*} [y \left (1\right ) = 1, y^{\prime }\left (1\right ) = 1] \end {align*}
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }-2 y^{3} y^{\prime } = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }-2 y^{3} y^{\prime }\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\frac {y^{4}}{2} = c_2 \end {align*}
Which is now solved for \(y\). 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 }&=\sqrt {y^{4}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {y^{4}+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 {1}{\sqrt {y^{4}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {y^{4}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Initial conditions are used to solve for the constants of integration.
Looking at the First solution \begin {align*} \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} = x +c_{2} \tag {1} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives \begin {align*} \int _{}^{1}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} = 1+c_{2}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = \sqrt {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} \right )+x +c_{2} \right )}^{4}+2 c_{1}} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = \sqrt {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} \right )+1+c_{2} \right )}^{4}+2 c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{2}\}\). There is no solution for the constants of integrations. This solution is removed.
Looking at the Second solution \begin {align*} \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} = x +c_{3} \tag {2} \end {align*}
Initial conditions are now substituted in the above solution. This will generate the required equations to solve for the integration constants. substituting \(y = 1\) and \(x = 1\) in the above gives \begin {align*} -\left (\int _{}^{1}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} \right ) = 1+c_{3}\tag {1A} \end {align*}
Taking derivative of the solution gives \begin {align*} y^{\prime } = -\sqrt {{\operatorname {RootOf}\left (-\left (\int _{}^{\textit {\_Z}}-\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} \right )+x +c_{3} \right )}^{4}+2 c_{1}} \end {align*}
substituting \(y^{\prime } = 1\) and \(x = 1\) in the above gives \begin {align*} 1 = -\sqrt {\operatorname {RootOf}\left (\int _{}^{\textit {\_Z}}\frac {1}{\sqrt {\textit {\_a}^{4}+2 c_{1}}}d \textit {\_a} +1+c_{3} \right )^{4}+2 c_{1}}\tag {2A} \end {align*}
Equations {1A,2A} are now solved for \(\{c_{1}, c_{3}\}\). There is no solution for the constants of integrations. This solution is removed.
Verification of solutions N/A
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 )-2 y^{3} = 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 {2 y^{3}}{p} \end {align*}
Where \(f(y)=2 y^{3}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= 2 y^{3} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {2 y^{3} \,d y} \\ \frac {p^{2}}{2}&=\frac {y^{4}}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {y^{4}}{2}-c_{1} = 0 \] Initial conditions are used to solve for \(c_{1}\). Substituting \(y=1\) and \(p=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -c_{1} = 0 \end {align*}
The solutions are \begin {align*} c_{1} = 0 \end {align*}
Trying the constant \begin {align*} c_{1} = 0 \end {align*}
Substituting \(c_{1}\) found above in the general solution gives \begin {align*} \frac {p^{2}}{2}-\frac {y^{4}}{2} = 0 \end {align*}
The constant \(c_{1} = 0\) gives valid solution.
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 }}^{2}}{2}-\frac {y^{4}}{2} = 0 \end {align*}
The ode \begin {align*} \frac {{y^{\prime }}^{2}}{2}-\frac {y^{4}}{2} = 0 \end {align*}
is factored to \begin {align*} \left (y^{2}-y^{\prime }\right ) \left (y^{2}+y^{\prime }\right ) = 0 \end {align*}
Which gives the following equations \begin {align*} y^{2}-y^{\prime } = 0\tag {1} \\ y^{2}+y^{\prime } = 0\tag {2} \\ \end {align*}
Each of the above equations is now solved.
Solving ODE (1) Integrating both sides gives \begin {align*} \int \frac {1}{y^{2}}d y &= \int {dx}\\ -\frac {1}{y}&= x +c_{2} \end {align*}
Initial conditions are used to solve for \(c_{2}\). Substituting \(x=1\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} -1 = 1+c_{2} \end {align*}
The solutions are \begin {align*} c_{2} = -2 \end {align*}
Trying the constant \begin {align*} c_{2} = -2 \end {align*}
Substituting \(c_{2}\) found above in the general solution gives \begin {align*} -\frac {1}{y} = x -2 \end {align*}
The constant \(c_{2} = -2\) gives valid solution.
Solving ODE (2) Integrating both sides gives \begin {align*} \int -\frac {1}{y^{2}}d y &= \int {dx}\\ \frac {1}{y}&= x +c_{3} \end {align*}
Initial conditions are used to solve for \(c_{3}\). Substituting \(x=1\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = 1+c_{3} \end {align*}
The solutions are \begin {align*} c_{3} = 0 \end {align*}
Trying the constant \begin {align*} c_{3} = 0 \end {align*}
Substituting \(c_{3}\) found above in the general solution gives \begin {align*} \frac {1}{y} = x \end {align*}
The constant \(c_{3} = 0\) gives valid solution.
Initial conditions are used to solve for the constants of integration.
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {1}{x -2} \\ \tag{2} y &= \frac {1}{x} \\ \end{align*}
Verification of solutions
\[ y = -\frac {1}{x -2} \] Verified OK.
\[ y = \frac {1}{x} \] Warning, solution could not be verified
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime \prime }-2 y^{3}=0, y \left (1\right )=1, y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & y^{\prime \prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} y^{\prime \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 ),y^{\prime \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 )-2 y^{3}=0 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=2 y^{3} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )d y =\int 2 y^{3}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=\frac {y^{4}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {y^{4}+2 c_{1}}, u \left (y \right )=-\sqrt {y^{4}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=\sqrt {y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y^{4}+2 c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y^{4}+2 c_{1}}}d x =\int 1d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2}\, \sqrt {4-\frac {2 \,\mathrm {I} y^{2} \sqrt {2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 \,\mathrm {I} y^{2} \sqrt {2}}{\sqrt {c_{1}}}}\, \mathit {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {\mathrm {I} \sqrt {2}}{\sqrt {c_{1}}}}}{2}, \mathrm {I}\right )}{4 \sqrt {\frac {\mathrm {I} \sqrt {2}}{\sqrt {c_{1}}}}\, \sqrt {y^{4}+2 c_{1}}}=x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}}, -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }=-\sqrt {y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sqrt {y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {y^{4}+2 c_{1}}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {y^{4}+2 c_{1}}}d x =\int \left (-1\right )d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {\sqrt {2}\, \sqrt {4-\frac {2 \,\mathrm {I} y^{2} \sqrt {2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 \,\mathrm {I} y^{2} \sqrt {2}}{\sqrt {c_{1}}}}\, \mathit {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {\mathrm {I} \sqrt {2}}{\sqrt {c_{1}}}}}{2}, \mathrm {I}\right )}{4 \sqrt {\frac {\mathrm {I} \sqrt {2}}{\sqrt {c_{1}}}}\, \sqrt {y^{4}+2 c_{1}}}=-x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & \left \{\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}}, -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}}\right \} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} \frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & \frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & -\frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-1+c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} \frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & \frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & \frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (1+c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & \frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-1+c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (-1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \\ \square & {} & \textrm {Check validity of solution}\hspace {3pt} -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use initial condition}\hspace {3pt} y \left (1\right )=1 \\ {} & {} & -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Compute derivative of the solution}\hspace {3pt} \\ {} & {} & -\frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Use the initial condition}\hspace {3pt} y^{\prime }{\raise{-0.36em}{\Big |}}{\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & -\frac {\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \mathrm {JacobiCN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (1+c_{2} \right ), \mathrm {I}\right ) \mathrm {JacobiDN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {2}}\, \left (1+c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \mathit {RootOf}\left (\textit {\_Z}^{2}-c_{1} , \mathit {index} =1\right )}{c_{1}}}} \\ {} & \circ & \textrm {Solve for}\hspace {3pt} c_{1} \hspace {3pt}\textrm {and}\hspace {3pt} c_{2} \\ {} & \circ & \textrm {The solution does not satisfy the initial conditions}\hspace {3pt} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville trying 2nd order WeierstrassP trying 2nd order JacobiSN <- 2nd_order JacobiSN successful`
✓ Solution by Maple
Time used: 0.047 (sec). Leaf size: 11
dsolve([diff(y(x),x$2)=2*y(x)^3,y(1) = 1, D(y)(1) = 1],y(x), singsol=all)
\[ y \left (x \right ) = -\frac {1}{x -2} \]
✓ Solution by Mathematica
Time used: 0.143 (sec). Leaf size: 12
DSolve[{y''[x]==2*y[x]^3,{y[1]==1,y'[1]==1}},y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {1}{2-x} \]