problem 52

Internal problem ID [7441]
Internal ﬁle name [OUTPUT/6408_Sunday_June_05_2022_04_46_57_PM_71426058/index.tex]

Book: Second order enumerated odes
Section: section 1
Problem number: 52.
ODE order: 2.
ODE degree: 3.

\[ \boxed {y {y^{\prime \prime }}^{3}+y^{3} {y^{\prime }}^{5}=0} \] The ode \begin {align*} y {y^{\prime \prime }}^{3}+y^{3} {y^{\prime }}^{5} = 0 \end {align*}

is factored to \begin {align*} y \left (y^{2} {y^{\prime }}^{5}+{y^{\prime \prime }}^{3}\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} y = 0\tag {1} \\ y^{2} {y^{\prime }}^{5}+{y^{\prime \prime }}^{3} = 0\tag {2} \\ \end {align*}

Solving ODE (1) Since $y = 0$, is missing derivative in $y$ then it is an algebraic equation. Solving for $y$. \begin {align*} \end {align*}

Solving ODE (2) 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*} y^{2} p \left (y \right )^{5}+p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end {align*}

Which is now solved as ﬁrst order ode for $p(y)$. The ode \begin {align*} y^{2} p \left (y \right )^{5}+p \left (y \right )^{3} \left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0 \end {align*}

is factored to \begin {align*} p \left (y \right )^{3} \left (p \left (y \right )^{2} y^{2}+\left (\frac {d}{d y}p \left (y \right )\right )^{3}\right ) = 0 \end {align*}

Which gives the following equations \begin {align*} p \left (y \right )^{3} = 0\tag {1} \\ p \left (y \right )^{2} y^{2}+\left (\frac {d}{d y}p \left (y \right )\right )^{3} = 0\tag {2} \\ \end {align*}

Solving ODE (1) Since $p \left (y \right )^{3} = 0$, is missing derivative in $p$ then it is an algebraic equation. Solving for $p \left (y \right )$. \begin {align*} \end {align*}

Solving ODE (2) The ode has the form \begin {align*} (p')^{\frac {n}{m}} &= f(y) g(p)\tag {1} \end {align*}

Where $n=3, m=1, f=-y^{2} , g=p^{2}$. Hence the ode is \begin {align*} (p')^{3} &= -p^{2} y^{2} \end {align*}

Solving for $\frac {d}{d y}p \left (y \right )$ from (1) gives \begin {align*} \frac {d}{d y}p \left (y \right ) &=\left (f g \right )^{\frac {1}{3}}\\ \frac {d}{d y}p \left (y \right ) &=-\frac {\left (f g \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (f g \right )^{\frac {1}{3}}}{2}\\ \frac {d}{d y}p \left (y \right ) &=-\frac {\left (f g \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (f g \right )^{\frac {1}{3}}}{2} \end {align*}

To be able to solve as separable ode, we have to now assume that $f>0,g>0$. \begin {align*} -y^{2} &> 0\\ p^{2} &> 0 \end {align*}

Under the above assumption the diﬀerential equations become separable and can be written as \begin {align*} \frac {d}{d y}p \left (y \right ) &=f^{\frac {1}{3}} g^{\frac {1}{3}}\\ \frac {d}{d y}p \left (y \right ) &=\frac {f^{\frac {1}{3}} g^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}{2}\\ \frac {d}{d y}p \left (y \right ) &=-\frac {f^{\frac {1}{3}} g^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}{2} \end {align*}

Therefore \begin {align*} \frac {1}{g^{\frac {1}{3}}} \, dp &= \left (f^{\frac {1}{3}}\right )\,dy\\ \frac {2}{g^{\frac {1}{3}} \left (i \sqrt {3}-1\right )} \, dp &= \left (f^{\frac {1}{3}}\right )\,dy\\ -\frac {2}{g^{\frac {1}{3}} \left (1+i \sqrt {3}\right )} \, dp &= \left (f^{\frac {1}{3}}\right )\,dy \end {align*}

Replacing $f(y),g(p)$ by their values gives \begin {align*} \frac {1}{\left (p^{2}\right )^{\frac {1}{3}}} \, dp &= \left (\left (-y^{2}\right )^{\frac {1}{3}}\right )\,dy\\ \frac {2}{\left (p^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )} \, dp &= \left (\left (-y^{2}\right )^{\frac {1}{3}}\right )\,dy\\ -\frac {2}{\left (p^{2}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )} \, dp &= \left (\left (-y^{2}\right )^{\frac {1}{3}}\right )\,dy \end {align*}

Integrating now gives the solutions. \begin {align*} \int \frac {1}{\left (p^{2}\right )^{\frac {1}{3}}}d p &= \int \left (-y^{2}\right )^{\frac {1}{3}}d y +c_{1}\\ \int \frac {2}{\left (p^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}d p &= \int \left (-y^{2}\right )^{\frac {1}{3}}d y +c_{1}\\ \int -\frac {2}{\left (p^{2}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d p &= \int \left (-y^{2}\right )^{\frac {1}{3}}d y +c_{1} \end {align*}

Integrating gives \begin {align*} \frac {3 p \left (y \right )}{\left (p \left (y \right )^{2}\right )^{\frac {1}{3}}} &= \frac {3 y \left (-y^{2}\right )^{\frac {1}{3}}}{5}+c_{1}\\ \frac {6 p \left (y \right )}{\left (p \left (y \right )^{2}\right )^{\frac {1}{3}} \left (i \sqrt {3}-1\right )} &= \frac {3 y \left (-y^{2}\right )^{\frac {1}{3}}}{5}+c_{1}\\ -\frac {6 p \left (y \right )}{\left (p \left (y \right )^{2}\right )^{\frac {1}{3}} \left (1+i \sqrt {3}\right )} &= \frac {3 y \left (-y^{2}\right )^{\frac {1}{3}}}{5}+c_{1} \end {align*}

Therefore \begin{align*} p \left (y \right ) &= -\frac {y^{5}}{125}+\frac {y^{2} \left (-y^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {y \left (-y^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27} \\ p \left (y \right ) &= -\frac {y^{5}}{125}+\frac {y^{2} \left (-y^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {y \left (-y^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27} \\ p \left (y \right ) &= -\frac {y^{5}}{125}+\frac {y^{2} \left (-y^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {y \left (-y^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27} \\ \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*} y^{\prime } = -\frac {y^{5}}{125}+\frac {y^{2} \left (-y^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {y \left (-y^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27} \end {align*}

Integrating both sides gives \begin {align*} \int _{}^{y}\frac {1}{-\frac {\textit {\_a}^{5}}{125}+\frac {\textit {\_a}^{2} \left (-\textit {\_a}^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {\textit {\_a} \left (-\textit {\_a}^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27}}d \textit {\_a} = x +c_{2} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{-\frac {\textit {\_a}^{5}}{125}+\frac {\textit {\_a}^{2} \left (-\textit {\_a}^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {\textit {\_a} \left (-\textit {\_a}^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27}}d \textit {\_a} &= x +c_{2} \\ \end{align*}

Veriﬁcation of solutions

\[ \int _{}^{y}\frac {1}{-\frac {\textit {\_a}^{5}}{125}+\frac {\textit {\_a}^{2} \left (-\textit {\_a}^{2}\right )^{\frac {2}{3}} c_{1}}{25}+\frac {\textit {\_a} \left (-\textit {\_a}^{2}\right )^{\frac {1}{3}} c_{1}^{2}}{15}+\frac {c_{1}^{3}}{27}}d \textit {\_a} = x +c_{2} \] Veriﬁed OK.

Summary

Veriﬁcation of solutions

Maple trace

`Methods for second order ODEs: 
   *** Sublevel 2 *** 
   Methods for second order ODEs: 
   Successful isolation of d^2y/dx^2: 3 solutions were found. Trying to solve each resulting ODE. 
      *** Sublevel 3 *** 
      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 
      -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)-(-_a^2*_b(_a)^2)^(1/3)*_b(_a) = 0, _b(_a), HINT = [[_a, 5*_b]]` 
         symmetry methods on request 
      `, `1st order, trying reduction of order with given symmetries:`[_a, 5*_b]

\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= c_{1} \\ \int _{}^{y \left (x \right )}\frac {1}{\operatorname {RootOf}\left (5 \left (\int _{\textit {\_g}}^{\textit {\_Z}}\frac {1}{\textit {\_a} \left (-\textit {\_f}^{2} \textit {\_a}^{2}\right )^{\frac {1}{3}}-5 \textit {\_f}}d \textit {\_f} \right )-\ln \left (\textit {\_a}^{5}+125\right )+5 c_{1} \right )}d \textit {\_a} -x -c_{2} &= 0 \\ \int _{}^{y \left (x \right )}\frac {1}{\operatorname {RootOf}\left (-i \ln \left (\textit {\_a}^{5}+125\right )+\sqrt {3}\, \ln \left (\textit {\_a}^{5}+125\right )+20 \left (\int _{\textit {\_g}}^{\textit {\_Z}}\frac {1}{2 i \textit {\_a} \left (-\textit {\_f}^{2} \textit {\_a}^{2}\right )^{\frac {1}{3}}+5 i \textit {\_f} +5 \sqrt {3}\, \textit {\_f}}d \textit {\_f} \right )-20 c_{1} \right )}d \textit {\_a} -x -c_{2} &= 0 \\ \int _{}^{y \left (x \right )}\frac {1}{\operatorname {RootOf}\left (20 \left (\int _{\textit {\_g}}^{\textit {\_Z}}\frac {1}{-2 i \textit {\_a} \left (-\textit {\_f}^{2} \textit {\_a}^{2}\right )^{\frac {1}{3}}-5 i \textit {\_f} +5 \sqrt {3}\, \textit {\_f}}d \textit {\_f} \right )+i \ln \left (\textit {\_a}^{5}+125\right )+\sqrt {3}\, \ln \left (\textit {\_a}^{5}+125\right )+20 c_{1} \right )}d \textit {\_a} -x -c_{2} &= 0 \\ \end{align*}

\begin{align*} y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},-\frac {3 i \left (-i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 i \left (i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \\ y(x)\to 0 \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 (-c_1)}\right )}{(-c_1){}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},-\frac {3 i \left (-i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 (-c_1)}\right )}{(-c_1){}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 i \left (i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 (-c_1)}\right )}{(-c_1){}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 \text {$\#$1}^{5/3}}{5 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},-\frac {3 i \left (-i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {27 \text {$\#$1} \operatorname {Hypergeometric2F1}\left (\frac {3}{5},3,\frac {8}{5},\frac {3 i \left (i+\sqrt {3}\right ) \text {$\#$1}^{5/3}}{10 c_1}\right )}{c_1{}^3}\&\right ][x+c_2] \\ \end{align*}

1.52 problem 52