Internal problem ID [10134]
Internal file name [OUTPUT/9081_Monday_June_06_2022_06_27_21_AM_73255612/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 6, non-linear second order
Problem number: 1813 (book 6.222).
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], _Liouville, [_2nd_order, _reducible, _mu_x_y1], [_2nd_order, _reducible, _mu_xy]]
\[ \boxed {y \left (1-\ln \left (y\right )\right ) y^{\prime \prime }+\left (1+\ln \left (y\right )\right ) {y^{\prime }}^{2}=0} \]
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*} \left (-\ln \left (y \right ) y +y \right ) p \left (y \right ) \left (\frac {d}{d y}p \left (y \right )\right )+\left (\ln \left (y \right ) p \left (y \right )+p \left (y \right )\right ) p \left (y \right ) = 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 {p \left (1+\ln \left (y \right )\right )}{y \left (-1+\ln \left (y \right )\right )} \end {align*}
Where \(f(y)=\frac {1+\ln \left (y \right )}{y \left (-1+\ln \left (y \right )\right )}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {1+\ln \left (y \right )}{y \left (-1+\ln \left (y \right )\right )} \,d y\\ \int { \frac {1}{p} \,dp} &= \int {\frac {1+\ln \left (y \right )}{y \left (-1+\ln \left (y \right )\right )} \,d y}\\ \ln \left (p \right )&=\ln \left (y \right )+2 \ln \left (-1+\ln \left (y \right )\right )+c_{1}\\ p&={\mathrm e}^{\ln \left (y \right )+2 \ln \left (-1+\ln \left (y \right )\right )+c_{1}}\\ &=c_{1} {\mathrm e}^{\ln \left (y \right )+2 \ln \left (-1+\ln \left (y \right )\right )} \end {align*}
Which simplifies to \[ p \left (y \right ) = c_{1} \left (y -2 \ln \left (y \right ) y +y \ln \left (y \right )^{2}\right ) \] 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 } = c_{1} \left (y-2 \ln \left (y\right ) y+y \ln \left (y\right )^{2}\right ) \end {align*}
Integrating both sides gives \begin {align*} \int \frac {1}{c_{1} y \left (1-2 \ln \left (y \right )+\ln \left (y \right )^{2}\right )}d y &= x +c_{2}\\ -\frac {1}{c_{1} \left (-1+\ln \left (y \right )\right )}&=x +c_{2} \end {align*}
Solving for \(y\) gives these solutions \begin {align*} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= {\mathrm e}^{\frac {c_{1} c_{2} +c_{1} x -1}{\left (x +c_{2} \right ) c_{1}}} \\ \end{align*}
Verification of solutions
\[ y = {\mathrm e}^{\frac {c_{1} c_{2} +c_{1} x -1}{\left (x +c_{2} \right ) c_{1}}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (-\ln \left (y\right ) y+y\right ) \left (\frac {d}{d x}y^{\prime }\right )+\left (\ln \left (y\right ) y^{\prime }+y^{\prime }\right ) y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Define new dependent variable}\hspace {3pt} u \\ {} & {} & u \left (x \right )=y^{\prime } \\ \bullet & {} & \textrm {Compute}\hspace {3pt} \frac {d}{d x}y^{\prime } \\ {} & {} & u^{\prime }\left (x \right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Use chain rule on the lhs}\hspace {3pt} \\ {} & {} & y^{\prime } \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Substitute in the definition of}\hspace {3pt} u \\ {} & {} & u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )=y^{\prime \prime } \\ \bullet & {} & \textrm {Make substitutions}\hspace {3pt} y^{\prime }=u \left (y \right ),\frac {d}{d x}y^{\prime }=u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )\hspace {3pt}\textrm {to reduce order of ODE}\hspace {3pt} \\ {} & {} & \left (-\ln \left (y \right ) y +y \right ) u \left (y \right ) \left (\frac {d}{d y}u \left (y \right )\right )+\left (\ln \left (y \right ) u \left (y \right )+u \left (y \right )\right ) u \left (y \right )=0 \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d y}u \left (y \right )=-\frac {\ln \left (y \right ) u \left (y \right )+u \left (y \right )}{-\ln \left (y \right ) y +y} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}=\frac {1+\ln \left (y \right )}{y \left (-1+\ln \left (y \right )\right )} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} y \\ {} & {} & \int \frac {\frac {d}{d y}u \left (y \right )}{u \left (y \right )}d y =\int \frac {1+\ln \left (y \right )}{y \left (-1+\ln \left (y \right )\right )}d y +c_{1} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & \ln \left (u \left (y \right )\right )=\ln \left (y \right )+2 \ln \left (-1+\ln \left (y \right )\right )+c_{1} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} y \left (-1+\ln \left (y \right )\right )^{2} \\ \bullet & {} & \textrm {Solve 1st ODE for}\hspace {3pt} u \left (y \right ) \\ {} & {} & u \left (y \right )={\mathrm e}^{c_{1}} y \left (-1+\ln \left (y \right )\right )^{2} \\ \bullet & {} & \textrm {Revert to original variables with substitution}\hspace {3pt} u \left (y \right )=y^{\prime },y =y \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} y \left (\ln \left (y\right )-1\right )^{2} \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }={\mathrm e}^{c_{1}} y \left (\ln \left (y\right )-1\right )^{2} \\ \bullet & {} & \textrm {Separate variables}\hspace {3pt} \\ {} & {} & \frac {y^{\prime }}{y \left (\ln \left (y\right )-1\right )^{2}}={\mathrm e}^{c_{1}} \\ \bullet & {} & \textrm {Integrate both sides with respect to}\hspace {3pt} x \\ {} & {} & \int \frac {y^{\prime }}{y \left (\ln \left (y\right )-1\right )^{2}}d x =\int {\mathrm e}^{c_{1}}d x +c_{2} \\ \bullet & {} & \textrm {Evaluate integral}\hspace {3pt} \\ {} & {} & -\frac {1}{\ln \left (y\right )-1}={\mathrm e}^{c_{1}} x +c_{2} \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} y \\ {} & {} & y={\mathrm e}^{\frac {{\mathrm e}^{c_{1}} x +c_{2} -1}{{\mathrm e}^{c_{1}} x +c_{2}}} \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying 2nd order Liouville <- 2nd_order Liouville successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 19
dsolve(y(x)*(1-ln(y(x)))*diff(diff(y(x),x),x)+(1+ln(y(x)))*diff(y(x),x)^2=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {c_{1} x +c_{2} -1}{c_{1} x +c_{2}}} \]
✓ Solution by Mathematica
Time used: 1.01 (sec). Leaf size: 34
DSolve[(1 + Log[y[x]])*y'[x]^2 + (1 - Log[y[x]])*y[x]*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to e^{\frac {c_1 x-1+c_2 c_1}{c_1 (x+c_2)}} \\ y(x)\to e \\ \end{align*}