Internal problem ID [4659]
Internal file name [OUTPUT/4152_Sunday_June_05_2022_12_29_48_PM_70386169/index.tex]
Book: Ordinary Differential Equations, By Tenenbaum and Pollard. Dover, NY 1963
Section: Chapter 8. Special second order equations. Lesson 35. Independent variable x
absent
Problem number: Exercise 35.9, page 504.
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]]
\[ \boxed {y^{\prime \prime }-2 k y^{3}=0} \]
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }-2 k y^{3} y^{\prime } = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }-2 k y^{3} y^{\prime }\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\frac {k 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 {k y^{4}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {k 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 {k \,y^{4}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {k \,y^{4}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} = x +c_{3} \] Verified OK.
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 k \,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 k \,y^{3}}{p} \end {align*}
Where \(f(y)=2 k \,y^{3}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= 2 k \,y^{3} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {2 k \,y^{3} \,d y} \\ \frac {p^{2}}{2}&=\frac {k \,y^{4}}{2}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {k \,y^{4}}{2}-c_{1} = 0 \] 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 {{y^{\prime }}^{2}}{2}-\frac {k y^{4}}{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 }&=\sqrt {k y^{4}+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {k 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 {k \,y^{4}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {1}{\sqrt {k \,y^{4}+2 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a}&= x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {\textit {\_a}^{4} k +2 c_{1}}}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime }-2 k y^{3}=0 \\ \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 k \,y^{3}=0 \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=2 k \,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 k \,y^{3}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \frac {u \left (y \right )^{2}}{2}=\frac {k \,y^{4}}{2}+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & \left \{u \left (y \right )=\sqrt {k \,y^{4}+2 c_{1}}, u \left (y \right )=-\sqrt {k \,y^{4}+2 c_{1}}\right \} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=\sqrt {k \,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 {k y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\sqrt {k y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {k y^{4}+2 c_{1}}}=1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {k 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} \sqrt {2}\, \sqrt {k}\, y^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 \,\mathrm {I} \sqrt {2}\, \sqrt {k}\, y^{2}}{\sqrt {c_{1}}}}\, \mathit {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}}{2}, \mathrm {I}\right )}{4 \sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}\, \sqrt {k 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 {k}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}}, -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {k}\, \sqrt {2}}\, \left (x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}}\right \} \\ \bullet & {} & \textrm {Solve 2nd ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )=-\sqrt {k \,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 {k y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\sqrt {k y^{4}+2 c_{1}} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{\sqrt {k y^{4}+2 c_{1}}}=-1 \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{\sqrt {k 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} \sqrt {2}\, \sqrt {k}\, y^{2}}{\sqrt {c_{1}}}}\, \sqrt {4+\frac {2 \,\mathrm {I} \sqrt {2}\, \sqrt {k}\, y^{2}}{\sqrt {c_{1}}}}\, \mathit {EllipticF}\left (\frac {y \sqrt {2}\, \sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}}{2}, \mathrm {I}\right )}{4 \sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}\, \sqrt {k 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 {k}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}}, -\frac {\mathrm {JacobiSN}\left (\sqrt {\mathrm {I} \sqrt {c_{1}}\, \sqrt {k}\, \sqrt {2}}\, \left (-x +c_{2} \right ), \mathrm {I}\right ) \sqrt {2}}{\sqrt {\frac {\mathrm {I} \sqrt {2}\, \sqrt {k}}{\sqrt {c_{1}}}}}\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 <- 2nd_order JacobiSN successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 20
dsolve(diff(y(x),x$2)=2*k*y(x)^3,y(x), singsol=all)
\[ y \left (x \right ) = c_{2} \operatorname {JacobiSN}\left (\left (\sqrt {-k}\, x +c_{1} \right ) c_{2} , i\right ) \]
✓ Solution by Mathematica
Time used: 61.304 (sec). Leaf size: 115
DSolve[y''[x]==2*k*y[x]^3,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {i \text {sn}\left (\left .(-1)^{3/4} \sqrt {\sqrt {k} \sqrt {c_1} (x+c_2){}^2}\right |-1\right )}{\sqrt {\frac {i \sqrt {k}}{\sqrt {c_1}}}} \\ y(x)\to \frac {i \text {sn}\left (\left .(-1)^{3/4} \sqrt {\sqrt {k} \sqrt {c_1} (x+c_2){}^2}\right |-1\right )}{\sqrt {\frac {i \sqrt {k}}{\sqrt {c_1}}}} \\ \end{align*}