2.2.6 problem 6

Internal problem ID [8805]
Book : Second order enumerated odes
Section : section 2
Problem number : 6
Date solved : Thursday, December 12, 2024 at 09:49:13 AM
CAS classification : [[_2nd_order, _missing_x], [_2nd_order, _reducible, _mu_x_y1]]

Solve

\begin{align*} y^{\prime \prime } y^{\prime }+y^{n}&=0 \end{align*}

Solved as second order missing x ode

Time used: 1.533 (sec)

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 \end{align*}

Then

\begin{align*} y'' &= \frac {dp}{dx}\\ &= \frac {dp}{dy}\frac {dy}{dx}\\ &= 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} = 0 \end{align*}

Which is now solved as first order ode for \(p(y)\).

The ode \(p^{\prime } = -\frac {y^{n}}{p^{2}}\) is separable as it can be written as

\begin{align*} p^{\prime }&= -\frac {y^{n}}{p^{2}}\\ &= f(y) g(p) \end{align*}

Where

\begin{align*} f(y) &= -y^{n}\\ g(p) &= \frac {1}{p^{2}} \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(p)} \,dp} &= \int { f(y) \,dy}\\ \int { p^{2}\,dp} &= \int { -y^{n} \,dy}\\ \frac {p^{3}}{3}&=-\frac {y^{n +1}}{n +1}+c_1 \end{align*}

Solving for \(p\) gives

\begin{align*} p &= \frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1} \\ p &= -\frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{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 )}^{{1}/{3}}}{2 \left (n +1\right )} \\ p &= -\frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{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 )}^{{1}/{3}}}{2 n +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*} y^{\prime } = \frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {n +1}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (c_1 n +c_1 \right )}{n +1}} \end{align*}

For solution (2) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\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 )}^{{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 )}^{{1}/{3}}}{2 \left (n +1\right )} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2 \left (n +1\right )}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (c_1 n +c_1 \right )}{n +1}} \end{align*}

For solution (3) found earlier, since \(p=y^{\prime }\) then we now have a new first order ode to solve which is

\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 )}^{{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 )}^{{1}/{3}}}{2 n +2} \end{align*}

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (\left (3 c_1 n -3 y^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}{2 n +2}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (c_1 n +c_1 \right )}{n +1}} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = {\mathrm e}^{\frac {\ln \left (c_1 n +c_1 \right )}{n +1}} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} \int _{}^{y}\frac {n +1}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau &= x +c_2 \\ \int _{}^{y}\frac {2 n +2}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau &= x +c_4 \\ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (3 c_1 n -3 \tau ^{n +1}+3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau &= x +c_3 \\ \end{align*}

Solved as second order can be made integrable

Time used: 1.503 (sec)

Multiplying the ode by \(y^{\prime }\) gives

\[ {y^{\prime }}^{2} y^{\prime \prime }+y^{n -1} y^{\prime } y = 0 \]

Integrating the above w.r.t \(x\) gives

\begin{align*} \int \left ({y^{\prime }}^{2} y^{\prime \prime }+y^{n -1} y^{\prime } y\right )d x &= 0 \\ \frac {{y^{\prime }}^{3}}{3}+\frac {y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}}{n +1} &= c_1 \end{align*}

Which is now solved for \(y\). Solving for the derivative gives these ODE’s to solve

\begin{align*} \tag{1} y^{\prime }&=\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1} \\ \tag{2} y^{\prime }&=-\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}-\frac {i \sqrt {3}\, {\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )} \\ \tag{3} y^{\prime }&=-\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 \left (n +1\right )}+\frac {i \sqrt {3}\, {\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y\right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{2 n +2} \\ \end{align*}

Now each of the above is solved separately.

Solving Eq. (1)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {n +1}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau = x +c_2 \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}{n +1}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (c_1 \left (n +1\right )\right )}{n +1}} \end{align*}

Solving Eq. (2)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau = x +c_3 \]

Singular solutions are found by solving

\begin{align*} -\frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}{2 \left (n +1\right )}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (c_1 \left (n +1\right )\right )}{n +1}} \end{align*}

Solving Eq. (3)

Unable to integrate (or intergal too complicated), and since no initial conditions are given, then the result can be written as

\[ \int _{}^{y}\frac {2 n +2}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau = x +c_4 \]

Singular solutions are found by solving

\begin{align*} \frac {{\left (\left (-3 y^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (y \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}{2 n +2}&= 0 \end{align*}

for \(y\). This is because we had to divide by this in the above step. This gives the following singular solution(s), which also have to satisfy the given ODE.

\begin{align*} y = {\mathrm e}^{\frac {\ln \left (c_1 \left (n +1\right )\right )}{n +1}} \end{align*}

Will add steps showing solving for IC soon.

The solution

\[ y = {\mathrm e}^{\frac {\ln \left (c_1 \left (n +1\right )\right )}{n +1}} \]

was found not to satisfy the ode or the IC. Hence it is removed.

Summary of solutions found

\begin{align*} \int _{}^{y}\frac {n +1}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \tau &= x +c_2 \\ \int _{}^{y}\frac {2 n +2}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (i \sqrt {3}-1\right )}d \tau &= x +c_4 \\ \int _{}^{y}-\frac {2 \left (n +1\right )}{{\left (\left (-3 \tau ^{2} {\mathrm e}^{\left (n -1\right ) \ln \left (\tau \right )}+3 c_1 n +3 c_1 \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}} \left (1+i \sqrt {3}\right )}d \tau &= x +c_3 \\ \end{align*}

Maple step by step solution

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*(n+1)/(n-2)]]`   *** 
   symmetry methods on request 
`, `1st order, trying reduction of order with given symmetries:`[-3/(n-2)*_a, -_b*(n+1)/(n-2)]
 

Maple dsolve solution

Solving time : 0.044 (sec)
Leaf size : 169

dsolve(diff(y(x),x)*diff(diff(y(x),x),x)+y(x)^n = 0, 
       y(x),singsol=all)
 

\begin{align*} \frac {\left (-2 n -2\right ) \left (\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} \right )-\left (x +c_{2} \right ) \left (1+i \sqrt {3}\right )}{1+i \sqrt {3}} &= 0 \\ -\frac {2 i \left (n +1\right ) \left (\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} \right )+\left (x +c_{2} \right ) \left (\sqrt {3}+i\right )}{\sqrt {3}+i} &= 0 \\ \left (\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} \right ) n +\int _{}^{y}\frac {1}{{\left (-\left (3 \textit {\_a}^{n +1}-c_{1} \right ) \left (n +1\right )^{2}\right )}^{{1}/{3}}}d \textit {\_a} -c_{2} -x &= 0 \\ \end{align*}

Mathematica DSolve solution

Solving time : 3.397 (sec)
Leaf size : 910

DSolve[{D[y[x],{x,2}]*D[y[x],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*}