2.6 problem 6

Internal problem ID [8095]
Book : Second order enumerated odes
Section : section 2
Problem number : 6
Date solved : Monday, October 21, 2024 at 04:49:31 PM
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*}

2.6.1 Solved as second order missing x ode

Time used: 1.532 (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 \(p\) from the above solution(s) gives (after possible removing of solutions that do not verify)

\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*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

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

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*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

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

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*}

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

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

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.

2.6.2 Solved as second order can be made integrable

Time used: 1.478 (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)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

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

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)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

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

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)

Since initial conditions \(\left (x_0,y_0\right ) \) are given, then the result can be written as Since unable to evaluate the integral, and no initial conditions are given, then the result becomes

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

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.

2.6.3 Maple step by step solution

2.6.4 Maple trace

Methods for second order ODEs:
 

2.6.5 Maple dsolve solution

Solving time : 0.016 (sec)
Leaf size : 169

dsolve(diff(diff(y(x),x),x)*diff(y(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*}

2.6.6 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*}