Internal problem ID [7106]
Internal file name [OUTPUT/6092_Sunday_June_05_2022_04_21_31_PM_77928761/index.tex]
Book: Own collection of miscellaneous problems
Section: section 1.0
Problem number: 62.
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]]
\[ \boxed {y y^{\prime \prime }=1} \]
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 p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right ) = 1 \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 {1}{y p} \end {align*}
Where \(f(y)=\frac {1}{y}\) and \(g(p)=\frac {1}{p}\). Integrating both sides gives \begin{align*} \frac {1}{\frac {1}{p}} \,dp &= \frac {1}{y} \,d y \\ \int { \frac {1}{\frac {1}{p}} \,dp} &= \int {\frac {1}{y} \,d y} \\ \frac {p^{2}}{2}&=\ln \left (y \right )+c_{1} \\ \end{align*} The solution is \[ \frac {p \left (y \right )^{2}}{2}-\ln \left (y \right )-c_{1} = 0 \] 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 }}^{2}}{2}-\ln \left (y\right )-c_{1} = 0 \end {align*}
Solving the given ode for \(y^{\prime }\) results in \(2\) differential equations to solve. Each one of these will generate a solution. The equations generated are \begin {align*} y^{\prime }&=\sqrt {2 \ln \left (y\right )+2 c_{1}} \tag {1} \\ y^{\prime }&=-\sqrt {2 \ln \left (y\right )+2 c_{1}} \tag {2} \end {align*}
Now each one of the above ODE is solved.
Solving equation (1)
Integrating both sides gives \begin {align*} \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )+2 c_{1}}}d \textit {\_a} = x +c_{2} \end {align*}
Solving equation (2)
Integrating both sides gives \begin {align*} \int _{}^{y}-\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )+2 c_{1}}}d \textit {\_a} = x +c_{3} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )+2 c_{1}}}d \textit {\_a} &= x +c_{2} \\ \tag{2} \int _{}^{y}-\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )+2 c_{1}}}d \textit {\_a} &= x +c_{3} \\ \end{align*}
Verification of solutions
\[ \int _{}^{y}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )+2 c_{1}}}d \textit {\_a} = x +c_{2} \] Verified OK.
\[ \int _{}^{y}-\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )+2 c_{1}}}d \textit {\_a} = x +c_{3} \] 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)-1/_a = 0, _b(_a), HINT = [[_a, 0]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, 0]
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 51
dsolve(y(x)*diff(y(x),x$2)=1,y(x), singsol=all)
\begin{align*} \int _{}^{y \left (x \right )}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )-c_{1}}}d \textit {\_a} -x -c_{2} &= 0 \\ -\left (\int _{}^{y \left (x \right )}\frac {1}{\sqrt {2 \ln \left (\textit {\_a} \right )-c_{1}}}d \textit {\_a} \right )-x -c_{2} &= 0 \\ \end{align*}
✓ Solution by Mathematica
Time used: 60.072 (sec). Leaf size: 93
DSolve[y[x]*y''[x]==1,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \exp \left (-\text {erf}^{-1}\left (-i \sqrt {\frac {2}{\pi }} \sqrt {e^{c_1} (x+c_2){}^2}\right ){}^2-\frac {c_1}{2}\right ) \\ y(x)\to \exp \left (-\text {erf}^{-1}\left (i \sqrt {\frac {2}{\pi }} \sqrt {e^{c_1} (x+c_2){}^2}\right ){}^2-\frac {c_1}{2}\right ) \\ \end{align*}