Internal problem ID [7447]
Internal file name [OUTPUT/6414_Sunday_June_05_2022_04_47_42_PM_82581/index.tex]
Book: Second order enumerated odes
Section: section 2
Problem number: 6.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_ode_missing_x"
Maple gives the following as the ode type
[[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]
\[ \boxed {y^{\prime } y^{\prime \prime }+y^{n}=0} \]
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 )^{2} \left (\frac {d}{d y}p \left (y \right )\right )+y^{n -1} y = 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 {y^{n -1} y}{p^{2}} \end {align*}
Where \(f(y)=-y^{n -1} y\) and \(g(p)=\frac {1}{p^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p^{2}}} \,dp &= -y^{n -1} y \,d y \\ \int { \frac {1}{\frac {1}{p^{2}}} \,dp} &= \int {-y^{n -1} y \,d y} \\ \frac {p^{3}}{3}&=-\frac {y^{n +1}}{n +1}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{3}}{3}+\frac {y^{n +1}}{n +1}-c_{1} = 0 \] 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 }}^{3}}{3}+\frac {y^{n +1}}{n +1}-c_{1} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(3\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\frac {{\left (\left (3 c_{1} n -3 y^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}{n +1} \tag {1} \\ y^{\prime }&=-\frac {{\left (\left (3 c_{1} n -3 y^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, {\left (\left (3 c_{1} n -3 y^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}{2 \left (n +1\right )} \tag {2} \\ y^{\prime }&=-\frac {{\left (\left (3 c_{1} n -3 y^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, {\left (\left (3 c_{1} n -3 y^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}{2 n +2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {n +1}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} = x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} = x +c_{3} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}d \textit {\_a} = x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {n +1}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} &= x +c_{3} \\ \tag{3} \int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}d \textit {\_a} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {n +1}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}} \left (1+i \sqrt {3}\right )}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_{1} n -3 \textit {\_a}^{n +1}+3 c_{1} \right ) \left (n +1\right )^{2}\right )}^{\frac {1}{3}} \left (i \sqrt {3}-1\right )}d \textit {\_a} = x +c_{4} \] 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 differential order: 2; missing variables `, `-> Computing symmetries using: way = 3 -> Calling odsolve with the ODE`, (diff(_b(_a), _a))*_b(_a)+_a^n/_b(_a) = 0, _b(_a), HINT = [[-3*_a/(n-2), -_b*(1+n)/(n-2)]]` *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[-3/(n-2)*_a, -_b*(1+n)/(n-2)]
✓ Solution by Maple
Time used: 0.031 (sec). Leaf size: 174
dsolve(diff(y(x),x$2)*diff(y(x),x)+y(x)^n=0,y(x), singsol=all)
\begin{align*} \frac {\left (-2-2 n \right ) \left (\int _{}^{y \left (x \right )}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_{1} \right ) \left (1+n \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} \right )-\left (1+i \sqrt {3}\right ) \left (x +c_{2} \right )}{1+i \sqrt {3}} &= 0 \\ -\frac {2 i \left (1+n \right ) \left (\int _{}^{y \left (x \right )}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_{1} \right ) \left (1+n \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} \right )+\left (x +c_{2} \right ) \left (\sqrt {3}+i\right )}{\sqrt {3}+i} &= 0 \\ \left (\int _{}^{y \left (x \right )}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_{1} \right ) \left (1+n \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} \right ) n +\int _{}^{y \left (x \right )}\frac {1}{{\left (-\left (3 \textit {\_a}^{1+n}-c_{1} \right ) \left (1+n \right )^{2}\right )}^{\frac {1}{3}}}d \textit {\_a} -c_{2} -x &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 2.4 (sec). Leaf size: 910
DSolve[y''[x]*y'[x]+y[x]^n==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt [3]{-\frac {1}{3}} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-\text {$\#$1}^{n+1}+c_1 (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{(-c_1) (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) (-c_1)}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 (-c_1) (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{(-c_1) (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) (-c_1)}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 (-c_1) (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt [3]{-\frac {1}{3}} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{(-c_1) (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) (-c_1)}\right )}{\sqrt [3]{-\text {$\#$1}^{n+1}+(-c_1) (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {\text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [\frac {(-1)^{2/3} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-3 \text {$\#$1}^{n+1}+3 c_1 (n+1)}}\&\right ][x+c_2] \\ y(x)\to \text {InverseFunction}\left [-\frac {\sqrt [3]{-\frac {1}{3}} \text {$\#$1} \sqrt [3]{n+1} \sqrt [3]{1-\frac {\text {$\#$1}^{n+1}}{c_1 (n+1)}} \operatorname {Hypergeometric2F1}\left (\frac {1}{3},\frac {1}{n+1},1+\frac {1}{n+1},\frac {\text {$\#$1}^{n+1}}{(n+1) c_1}\right )}{\sqrt [3]{-\text {$\#$1}^{n+1}+c_1 (n+1)}}\&\right ][x+c_2] \\ \end{align*}