Internal problem ID [12203]
Internal file name [OUTPUT/10856_Thursday_September_21_2023_05_48_08_AM_39220548/index.tex]
Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS,
MOSCOW, Third printing 1977.
Section: Chapter 2, DIFFERENTIAL EQUATIONS OF THE SECOND ORDER AND HIGHER.
Problems page 172
Problem number: Problem 56.
ODE order: 2.
ODE degree: 0.
The type(s) of ODE detected by this program : "second_order_ode_missing_y"
Maple gives the following as the ode type
[[_2nd_order, _missing_y]]
\[ \boxed {x y^{\prime \prime }-y^{\prime } \ln \left (\frac {y^{\prime }}{x}\right )=0} \]
This is second order ode with missing dependent variable \(y\). Let \begin {align*} p(x) &= y^{\prime } \end {align*}
Then \begin {align*} p'(x) &= y^{\prime \prime } \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (x \right ) x -p \left (x \right ) \ln \left (\frac {p \left (x \right )}{x}\right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. Using the change of variables \(p \left (x \right ) = u \left (x \right ) x\) on the above ode results in new ode in \(u \left (x \right )\) \begin {align*} \left (u^{\prime }\left (x \right ) x +u \left (x \right )\right ) x -u \left (x \right ) x \ln \left (u \left (x \right )\right ) = 0 \end {align*}
In canonical form the ODE is \begin {align*} u' &= F(x,u)\\ &= f( x) g(u)\\ &= \frac {u \left (\ln \left (u \right )-1\right )}{x} \end {align*}
Where \(f(x)=\frac {1}{x}\) and \(g(u)=u \left (\ln \left (u \right )-1\right )\). Integrating both sides gives \begin{align*} \frac {1}{u \left (\ln \left (u \right )-1\right )} \,du &= \frac {1}{x} \,d x \\ \int { \frac {1}{u \left (\ln \left (u \right )-1\right )} \,du} &= \int {\frac {1}{x} \,d x} \\ \ln \left (\ln \left (u \right )-1\right )&=\ln \left (x \right )+c_{2} \\ \end{align*} Raising both side to exponential gives \begin {align*} \ln \left (u \right )-1 &= {\mathrm e}^{\ln \left (x \right )+c_{2}} \end {align*}
Which simplifies to \begin {align*} \ln \left (u \right )-1 &= c_{3} x \end {align*}
Therefore the solution \(p \left (x \right )\) is \begin {align*} p \left (x \right )&=x u\\ &=x \,{\mathrm e}^{1+c_{3} {\mathrm e}^{c_{2}} x} \end {align*}
Since \(p=y^{\prime }\) then the new first order ode to solve is \begin {align*} y^{\prime } = x \,{\mathrm e}^{1+c_{3} {\mathrm e}^{c_{2}} x} \end {align*}
Integrating both sides gives \begin {align*} y &= \int { x \,{\mathrm e}^{1+c_{3} {\mathrm e}^{c_{2}} x}\,\mathop {\mathrm {d}x}}\\ &= \frac {\left (c_{3} {\mathrm e}^{c_{2}} x -1\right ) {\mathrm e}^{1+c_{3} {\mathrm e}^{c_{2}} x -2 c_{2}}}{c_{3}^{2}}+c_{4} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} {\mathrm e}^{c_{2}} x -1\right ) {\mathrm e}^{1+c_{3} {\mathrm e}^{c_{2}} x -2 c_{2}}}{c_{3}^{2}}+c_{4} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} {\mathrm e}^{c_{2}} x -1\right ) {\mathrm e}^{1+c_{3} {\mathrm e}^{c_{2}} x -2 c_{2}}}{c_{3}^{2}}+c_{4} \] 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 -> Calling odsolve with the ODE`, diff(_b(_a), _a) = _b(_a)*ln(_b(_a)/_a)/_a, _b(_a), HINT = [[_a, _b]]` *** Sublevel 2 *** symmetry methods on request `, `1st order, trying reduction of order with given symmetries:`[_a, _b]
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 31
dsolve(x*diff(y(x),x$2)=diff(y(x),x)*ln(diff(y(x),x)/x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {{\mathrm e}^{c_{1} x +1} c_{1} x +c_{2} c_{1}^{2}-{\mathrm e}^{c_{1} x +1}}{c_{1}^{2}} \]
✓ Solution by Mathematica
Time used: 0.905 (sec). Leaf size: 31
DSolve[x*y''[x]==y'[x]*Log[y'[x]/x],y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{e^{c_1} x+1-2 c_1} \left (-1+e^{c_1} x\right )+c_2 \]