Internal problem ID [9919]
Internal file name [OUTPUT/8866_Monday_June_06_2022_05_42_24_AM_99880625/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1597 (6.7).
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]]
\[ \boxed {y^{\prime \prime }-a y^{3}=0} \]
Multiplying the ode by \(y^{\prime }\) gives \[ y^{\prime } y^{\prime \prime }-a y^{\prime } y^{3} = 0 \] Integrating the above w.r.t \(x\) gives \begin {align*} \int \left (y^{\prime } y^{\prime \prime }-a y^{\prime } y^{3}\right )d x &= 0 \\ \frac {{y^{\prime }}^{2}}{2}-\frac {a y^{4}}{4} = 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 }&=\frac {\sqrt {2 a y^{4}+8 c_{1}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2 a y^{4}+8 c_{1}}}{2} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {2}{\sqrt {2 a \,y^{4}+8 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {2}{\sqrt {2 a \,y^{4}+8 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a}&= c_{3} +x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} &= c_{3} +x \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} = c_{3} +x \] 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 )-a \,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 {a \,y^{3}}{p} \end {align*}
Where \(f(y)=a \,y^{3}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= a \,y^{3} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {a \,y^{3} \,d y} \\ \frac {p^{2}}{2}&=\frac {a \,y^{4}}{4}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\frac {a \,y^{4}}{4}-c_{1} = 0 \] 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 {a y^{4}}{4}-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 {2 a y^{4}+8 c_{1}}}{2} \tag {1} \\ y^{\prime }&=-\frac {\sqrt {2 a y^{4}+8 c_{1}}}{2} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {2}{\sqrt {2 a \,y^{4}+8 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int -\frac {2}{\sqrt {2 a \,y^{4}+8 c_{1}}}d y &= \int {dx}\\ \int _{}^{y}-\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a}&= c_{3} +x \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} &= c_{3} +x \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {2}{\sqrt {2 \textit {\_a}^{4} a +8 c_{1}}}d \textit {\_a} = c_{3} +x \] Verified OK.
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: 0.453 (sec). Leaf size: 26
dsolve(diff(diff(y(x),x),x)-a*y(x)^3=0,y(x), singsol=all)
\[ y \left (x \right ) = c_{2} \operatorname {JacobiSN}\left (\frac {\left (\sqrt {2}\, \sqrt {-a}\, x +2 c_{1} \right ) c_{2}}{2}, i\right ) \]
✓ Solution by Mathematica
Time used: 61.747 (sec). Leaf size: 131
DSolve[-(a*y[x]^3) + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {i \sqrt [4]{2} \text {sn}\left (\left .-\frac {(1-i) \sqrt {\sqrt {a} \sqrt {c_1} (x+c_2){}^2}}{2^{3/4}}\right |-1\right )}{\sqrt {\frac {i \sqrt {a}}{\sqrt {c_1}}}} \\ y(x)\to \frac {i \sqrt [4]{2} \text {sn}\left (\left .-\frac {(1-i) \sqrt {\sqrt {a} \sqrt {c_1} (x+c_2){}^2}}{2^{3/4}}\right |-1\right )}{\sqrt {\frac {i \sqrt {a}}{\sqrt {c_1}}}} \\ \end{align*}