problem 1753 (book 6.162)

Internal problem ID [10074]
Internal ﬁle name [OUTPUT/9021_Monday_June_06_2022_06_14_04_AM_88608185/index.tex]

Book: Diﬀerential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1753 (book 6.162).
ODE order: 2.
ODE degree: 1.

7.162.1 Solving as second order ode missing x ode

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*} 4 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) y -5 p \left (y \right )^{2}+a \,y^{2} = 0 \end {align*}

Which is now solved as ﬁrst order ode for \(p(y)\). Using the change of variables \(p \left (y \right ) = u \left (y \right ) y\) on the above ode results in new ode in \(u \left (y \right )\) \begin {align*} 4 u \left (y \right ) y^{2} \left (\left (\frac {d}{d y}u \left (y \right )\right ) y +u \left (y \right )\right )-5 u \left (y \right )^{2} y^{2} = -a \,y^{2} \end {align*}

In canonical form the ODE is \begin {align*} u' &= F(y,u)\\ &= f( y) g(u)\\ &= \frac {u^{2}-a}{4 u y} \end {align*}

Where \(f(y)=\frac {1}{4 y}\) and \(g(u)=\frac {u^{2}-a}{u}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {u^{2}-a}{u}} \,du &= \frac {1}{4 y} \,d y \\ \int { \frac {1}{\frac {u^{2}-a}{u}} \,du} &= \int {\frac {1}{4 y} \,d y} \\ \frac {\ln \left (u^{2}-a \right )}{2}&=\frac {\ln \left (y \right )}{4}+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \sqrt {u^{2}-a} &= {\mathrm e}^{\frac {\ln \left (y \right )}{4}+c_{2}} \end {align*}

Which simpliﬁes to \begin {align*} \sqrt {u^{2}-a} &= c_{3} y^{\frac {1}{4}} \end {align*}

Which simpliﬁes to \[ \sqrt {u \left (y \right )^{2}-a} = c_{3} y^{\frac {1}{4}} {\mathrm e}^{c_{2}} \] The solution is \[ \sqrt {u \left (y \right )^{2}-a} = c_{3} y^{\frac {1}{4}} {\mathrm e}^{c_{2}} \] Replacing \(u(y)\) in the above solution by \(\frac {p \left (y \right )}{y}\) results in the solution for \(p \left (y \right )\) in implicit form \begin {align*} \sqrt {\frac {p \left (y \right )^{2}}{y^{2}}-a} = c_{3} y^{\frac {1}{4}} {\mathrm e}^{c_{2}}\\ \sqrt {\frac {-a \,y^{2}+p \left (y \right )^{2}}{y^{2}}} = c_{3} y^{\frac {1}{4}} {\mathrm e}^{c_{2}} \end {align*}

For solution (1) found earlier, since \(p=y^{\prime }\) then we now have a new ﬁrst order ode to solve which is \begin {align*} \sqrt {\frac {-a y^{2}+{y^{\prime }}^{2}}{y^{2}}} = c_{3} y^{\frac {1}{4}} {\mathrm e}^{c_{2}} \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 {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}} \tag {1} \\ y^{\prime }&=-\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}} \tag {2} \end {align*}

Integrating both sides gives \begin{align*} \int \frac {1}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a \,y^{2}}}d y &= \int d x \\ -\frac {4 y \sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}}{\sqrt {a}}\right )}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}}\, \sqrt {a}}&=x +c_{4} \\ \end{align*} Solving equation (2)

Integrating both sides gives \begin{align*} \int -\frac {1}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a \,y^{2}}}d y &= \int d x \\ \frac {4 y \sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}}{\sqrt {a}}\right )}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}}\, \sqrt {a}}&=x +c_{5} \\ \end{align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} -\frac {4 y \sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}}{\sqrt {a}}\right )}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}}\, \sqrt {a}} &= x +c_{4} \\ \tag{2} \frac {4 y \sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}}{\sqrt {a}}\right )}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}}\, \sqrt {a}} &= x +c_{5} \\ \end{align*}

Veriﬁcation of solutions

\[ -\frac {4 y \sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}}{\sqrt {a}}\right )}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}}\, \sqrt {a}} = x +c_{4} \] Veriﬁed OK.

\[ \frac {4 y \sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}\, \operatorname {arctanh}\left (\frac {\sqrt {{\mathrm e}^{2 c_{2}} c_{3}^{2} \sqrt {y}+a}}{\sqrt {a}}\right )}{\sqrt {y^{\frac {5}{2}} {\mathrm e}^{2 c_{2}} c_{3}^{2}+a y^{2}}\, \sqrt {a}} = x +c_{5} \] Veriﬁed OK.

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 a quadrature 
checking if the LODE has constant coefficients 
<- constant coefficients successful 
<- 2nd order, 2 integrating factors of the form mu(x,y) successful`

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= \frac {16 \,{\mathrm e}^{\sqrt {a}\, x} a^{2}}{\left ({\mathrm e}^{\frac {\sqrt {a}\, x}{2}} c_{1} -c_{2} \right )^{4}} \\ \end{align*}

7.162 problem 1753 (book 6.162)

7.162.1 Solving as second order ode missing x ode