Internal problem ID [7446]
Internal file name [OUTPUT/6413_Sunday_June_05_2022_04_47_37_PM_44932305/index.tex]
Book: Second order enumerated odes
Section: section 2
Problem number: 5.
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]]
\[ \boxed {y^{\prime \prime } y^{\prime }+y^{2}=0} \]
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^{2} = 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^{2}}{p^{2}} \end {align*}
Where \(f(y)=-y^{2}\) and \(g(p)=\frac {1}{p^{2}}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p^{2}}} \,dp &= -y^{2} \,d y \\ \int { \frac {1}{\frac {1}{p^{2}}} \,dp} &= \int {-y^{2} \,d y} \\ \frac {p^{3}}{3}&=-\frac {y^{3}}{3}+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{3}}{3}+\frac {y^{3}}{3}-c_{1} = 0 \] 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^{3}}{3}-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 }&=\left (-y^{3}+3 c_{1} \right )^{\frac {1}{3}} \tag {1} \\ y^{\prime }&=-\frac {\left (-y^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (-y^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2} \tag {2} \\ y^{\prime }&=-\frac {\left (-y^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (-y^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2} \tag {3} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int \frac {1}{\left (-y^{3}+3 c_{1} \right )^{\frac {1}{3}}}d y &= \int {dx}\\ \int _{}^{y}\frac {1}{\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}d \textit {\_a}&= x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {1}{-\frac {\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}}d \textit {\_a} = x +c_{3} \end {align*}
Solving equation (3)
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {1}{-\frac {\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}}d \textit {\_a} = x +c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}\frac {1}{-\frac {\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}}d \textit {\_a} &= x +c_{3} \\ \tag{3} \int _{}^{y}\frac {1}{-\frac {\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}}d \textit {\_a} &= x +c_{4} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}\frac {1}{-\frac {\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}-\frac {i \sqrt {3}\, \left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}}d \textit {\_a} = x +c_{3} \] Verified OK.
\[ \int _{}^{y}\frac {1}{-\frac {\left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}+\frac {i \sqrt {3}\, \left (-\textit {\_a}^{3}+3 c_{1} \right )^{\frac {1}{3}}}{2}}d \textit {\_a} = x +c_{4} \] 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 Try integration with the canonical coordinates of the symmetry [0, y] -> Calling odsolve with the ODE`, diff(_b(_a), _a) = -(_b(_a)^3+1)/_b(_a), _b(_a), explicit, HINT = [[1, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[1, 0]
✓ Solution by Maple
Time used: 0.078 (sec). Leaf size: 61
dsolve(diff(y(x),x$2)*diff(y(x),x)+y(x)^2=0,y(x), singsol=all)
\begin{align*} y \left (x \right ) &= 0 \\ y \left (x \right ) &= {\mathrm e}^{\frac {\sqrt {3}\, \left (\int \tan \left (\operatorname {RootOf}\left (-\sqrt {3}\, \ln \left (\cos \left (\textit {\_Z} \right )^{2}\right )-2 \sqrt {3}\, \ln \left (\tan \left (\textit {\_Z} \right )+\sqrt {3}\right )+6 \sqrt {3}\, c_{1} +6 \sqrt {3}\, x +6 \textit {\_Z} \right )\right )d x \right )}{2}+c_{2} +\frac {x}{2}} \\ \end{align*}
✓ Solution by Mathematica
Time used: 1.356 (sec). Leaf size: 180
DSolve[y''[x]*y'[x]+y[x]^2==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_2}{\sqrt [3]{1+\text {InverseFunction}\left [\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][-x+c_1]} \sqrt [3]{\text {InverseFunction}\left [\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][-x+c_1]{}^2-\text {InverseFunction}\left [\frac {1}{6} \log \left (\text {$\#$1}^2-\text {$\#$1}+1\right )+\frac {\arctan \left (\frac {2 \text {$\#$1}-1}{\sqrt {3}}\right )}{\sqrt {3}}-\frac {1}{3} \log (\text {$\#$1}+1)\&\right ][-x+c_1]+1}} \]